This deck covers everything a parabola can tell you: vertex form and the sign trap hiding inside it, the vertex formula from standard form, converting between forms by completing the square, intercepts and what the discriminant says about how many there are, domain and range, and applied optimization for maximum area, maximum revenue, and projectile height. It targets the four errors that cost the most points: flipping the sign of the vertex, dropping the negative in the vertex formula, reporting where the maximum happens instead of the maximum value, and assuming that every parabola crosses the axis twice.
Subject: College Algebra · 136 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Title
College Algebra - Deck 12
Vertex form, the vertex formula, completing the square, intercepts and the discriminant, and the optimization problems they all exist to solve.
Objectives
A parabola is the second shape you meet in this course, right after the line. Almost every question about one comes down to a single point: the vertex.
Warm-up
Discussion prompt
Before we open Quadratic Functions, Parabolas, and Optimization: without looking back, what was the main idea of Linear Functions and Modeling, and what could you do by the end of it that you could not do before?
Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.
Answer:
That deck covers everything linear. It shows a constant rate of change in a table, in a graph, and in an equation, finds slope from two points and reads it as a rate with units, and works through slope-intercept, point-slope, and standard form, along with horizontal and vertical lines and the slopes of parallel and perpendicular lines. It then builds a model from a description or from two data points and predicts with it, and closes with scatter plots, best-fit lines, and correlation. It targets four classic errors: subtracting the coordinates in mismatched order, confusing zero slope with undefined slope, taking only half of a negative reciprocal, and reading the y-intercept off an equation that has not yet been solved for y.
Section
Part 1
Concept
A quadratic function is a polynomial function of degree two. That is the whole definition: the highest power on the variable is two.
\[ f(x) = ax^2 + bx + c, \qquad a \ne 0 \]
The condition on the leading coefficient matters. If it were zero the squared term would vanish and you would be left with a line, not a parabola.
| Function | Quadratic? | Why |
|---|---|---|
| f(x) = 3x^2 - 5x + 1 | yes | degree two, leading coefficient 3 |
| f(x) = 7 - x^2 | yes | degree two; a = -1, b = 0, c = 7 |
| f(x) = 4x + 9 | no | degree one - a line |
| f(x) = x^3 - x^2 | no | degree three |
Discrimination
Sort into buckets
Sort these by Quadratic?, from memory, without looking back at A quadratic function has a squared term and…. Telling them apart on the spot is the skill; the table is only where the answer happens to be written down.
Picture it
Animation
Shows: Solving a quadratic inequality — a rendered Manim animation.
Rendered with Manim.
Takeaway: Below the axis between the roots — read the answer off the picture.
Intuition
A line has one constant rate of change, so it never turns. A quadratic has a rate of change that is itself changing at a steady pace - and that is what bends the graph into a curve.
Watch the squaring function on the whole numbers. Look at the last column.
| input | output | change from the row above | change in the change |
|---|---|---|---|
| 0 | 0 | - | - |
| 1 | 1 | 1 | - |
| 2 | 4 | 3 | 2 |
| 3 | 9 | 5 | 2 |
| 4 | 16 | 7 | 2 |
The outputs climb faster and faster, but the speed-up itself is constant. That steady acceleration is the signature of a quadratic - and it is why the curve, coming down, must eventually stop and go back up.
parabola — The U-shaped curve that is the graph of every quadratic function. It is symmetric, it turns exactly once, and it never straightens out.
Pattern
Step through it
Step through Why the graph has to turn around one row at a time. What is driving the change, and what would the row after the last one be?
Picture it
Figure (svg): A parabola opening upward with a dashed vertical axis of symmetry through its lowest point, a marked vertex, and two marked points where the curve crosses the horizontal axis.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
The axis of symmetry is a line, so its equation always looks like the variable set equal to a number - never a single number by itself.
Concept
Figure (svg): A parabola opening upward with a dashed vertical axis of symmetry through its lowest point, a marked vertex, and two marked points where the curve crosses the horizontal axis.
vertex — The single turning point of the parabola - the lowest point if it opens up, the highest point if it opens down. Everything in this deck is really about finding it.
axis of symmetry — The vertical line through the vertex. Fold the graph along it and the two halves land exactly on each other.
The axis of symmetry is a line, so its equation always looks like the variable set equal to a number - never a single number by itself.
Definition probe
Sort into buckets
Every line below is part of the definition of parabola or of vertex — one or the other, never both. Put each where it belongs.
Picture it
Figure (svg): Two parabolas side by side: the left one opens upward with its vertex at the bottom, the right one opens downward with its vertex at the top.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
If the leading coefficient is positive, the parabola opens up and the vertex is the lowest point - a minimum.
Concept
Figure (svg): Two parabolas side by side: the left one opens upward with its vertex at the bottom, the right one opens downward with its vertex at the top.
If the leading coefficient is positive, the parabola opens up and the vertex is the lowest point - a minimum.
\[ a > 0 \;\Longrightarrow\; \text{opens up, vertex is a minimum} \]
If it is negative, the parabola opens down and the vertex is the highest point - a maximum.
\[ a < 0 \;\Longrightarrow\; \text{opens down, vertex is a maximum} \]
This one sign decides whether an application is asking for a largest value or a smallest one. Check it first, every time.
Counterexample
Discussion prompt
If the leading coefficient is positive, the parabola opens up and the vertex is the lowest point - a minimum.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
If it is negative, the parabola opens down and the vertex is the highest point - a maximum.
Picture it
Animation
Shows: The sign of a decides max or min — a rendered Manim animation.
Rendered with Manim.
Takeaway: Opening up gives a minimum; opening down gives a maximum.
Intuition
Think of the leading coefficient as a dial that stretches the basic squaring curve vertically. Bigger magnitude, steeper climb, and the parabola looks narrower.
| Function | Output at one unit from the vertex | Look |
|---|---|---|
| f(x) = x^2 | 1 | the parent shape |
| f(x) = 3x^2 | 3 | three times as tall, so narrower |
| f(x) = 0.25x^2 | 0.25 | a quarter as tall, so wider |
Nothing here moves the curve sideways or up and down. A stretch pins the vertex in place and pulls the arms.
Comparison
Comparison matrix
From The leading coefficient is also a width dial: refill the Output at one unit from the vertex column from what you know. The rest of the table is as it appeared.
| Function | Output at one unit from the vertex | Look |
|---|---|---|
| f(x) = x^2 | 1 | the parent shape |
| f(x) = 3x^2 | 3 | three times as tall, so narrower |
| f(x) = 0.25x^2 | 0.25 | a quarter as tall, so wider |
Section
Part 2
Concept
There is a way to write a quadratic that puts the vertex right on the page, no work required.
\[ f(x) = a(x - h)^2 + k \]
The vertex is the point whose coordinates are the two numbers in that expression, in the order they appear.
\[ \text{vertex } (h,\ k), \qquad \text{axis of symmetry } x = h \]
Same leading coefficient as before: it still sets the opening direction and the stretch. Vertex form changes the packaging, not the parabola.
Analogy
Discussion prompt
Explain Vertex form hands you the vertex by analogy to something with no College Algebra in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
The vertex is the point whose coordinates are the two numbers in that expression, in the order they appear.
Intuition
A square of a real number is never negative. The smallest it can possibly be is zero, and it is zero exactly when the thing being squared is zero.
\[ (x - h)^2 \ge 0, \quad \text{with equality only when } x = h \]
So if the leading coefficient is positive, the whole term adds nothing at that one input and adds something positive everywhere else. The output bottoms out right there.
\[ f(h) = a\cdot 0 + k = k \]
That is the entire reason the vertex is where it is. You are not memorizing a rule - you are reading off the one input that switches the squared term off.
Explain it
Discussion prompt
Explain Why that form works: the squared part is never negative to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
A square of a real number is never negative. The smallest it can possibly be is zero, and it is zero exactly when the thing being squared is zero.
Concept
\[ f(x) = -3(x - 4)^2 + 10 \]
\[ \text{vertex } (4,\ 10), \qquad x = 4, \qquad \text{maximum value } 10 \]
No graphing, no algebra. That is why vertex form is worth converting to.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Reading the number inside the parentheses straight off the page.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The plus three is sitting right there, so it feels like the horizontal coordinate.
The form has a minus built into it. Rewrite until you literally see a minus sign.
Why: The plus three is sitting right there, so it feels like the horizontal coordinate.
Trap
Reading the number inside the parentheses straight off the page.
\[ f(x) = (x + 3)^2 - 5 \]
Claim the vertex is at three to the right and five down
Why: The plus three is sitting right there, so it feels like the horizontal coordinate.
Test it: the claimed vertex output does not match
Why: Substituting three gives thirty-six minus five, which is thirty-one - nowhere near the minimum of this upward parabola.
\[ f(3) = (3+3)^2 - 5 = 36 - 5 = 31 \]
The form has a minus built into it. Rewrite until you literally see a minus sign.
\[ f(x) = a(x - h)^2 + k \]
Rewrite the plus three as minus a negative three
Why: Now the expression matches the template exactly, so the number after the minus sign is the horizontal coordinate.
\[ f(x) = \bigl(x - (-3)\bigr)^2 - 5 \]
Verify by substituting negative three
Why: The squared term switches off and the output is exactly the constant, which is the minimum for an upward parabola.
\[ f(-3) = (0)^2 - 5 = -5 \quad \Rightarrow \quad \text{vertex } (-3,\ -5) \]
Notation
Annotate
From Trap: the vertex is not the number you see — read this one piece at a time. What is each part doing?
On: \( f(x) = (x + 3)^2 - 5 \)
Ranking
Put in order
Put the moves of Worked example: a full graph from vertex form into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The form already matches the template with a minus sign inside, so the coordinates are three and negative eight.
Worked example
Give the vertex, the opening direction, the axis of symmetry, and both kinds of intercept.
\[ f(x) = 2(x - 3)^2 - 8 \]
Read the vertex and the opening direction
Why: The form already matches the template with a minus sign inside, so the coordinates are three and negative eight. The leading coefficient is positive, so it opens up and that vertex is a minimum.
\[ \text{vertex } (3,\ -8), \qquad \text{opens up} \]
Write the axis of symmetry as an equation of a line
Why: The mirror line is vertical through the vertex, so it is the variable set equal to the horizontal coordinate.
\[ x = 3 \]
Find the vertical intercept by substituting zero
Why: The graph meets the vertical axis where the input is zero; substitute and simplify.
\[ f(0) = 2(0-3)^2 - 8 = 2(9) - 8 = 10 \]
Find the horizontal intercepts by setting the output to zero
Why: Vertex form is already isolated, so the square root property finishes it in two moves - and the plus-or-minus is what gives both crossings.
\[ 2(x-3)^2 - 8 = 0 \;\Rightarrow\; (x-3)^2 = 4 \;\Rightarrow\; x - 3 = \pm 2 \]
\[ x = 5 \quad \text{or} \quad x = 1 \]
Verify both crossings in the original function
Why: Substituting each candidate must return zero, and the two crossings must sit the same distance from the axis of symmetry - they do, both two units away.
\[ f(5) = 2(2)^2 - 8 = 0, \qquad f(1) = 2(-2)^2 - 8 = 0 \]
Picture it
Animation
Shows: Each line of the worked example "a full graph from vertex form", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Substituting each candidate must return zero, and the two crossings must sit the same distance from the axis of symmetry - they do, both two units away.
Pattern
Whenever a quadratic is already written in this shape, you never have to compute anything to describe its graph.
\[ f(x) = a(x - h)^2 + k \]
One safety habit finishes it: substitute the first coordinate back into the rule. The squared term should switch off and leave exactly the second coordinate.
Check
Rewrite it until you can see a minus sign inside the parentheses, then answer.
\[ f(x) = -4(x + 7)^2 + 2 \]
Check your understanding
What is the vertex of this parabola?
Answer: A
Why: Vertex form subtracts h inside the parentheses, so x plus 7 must be rewritten as x minus negative 7, making h equal to negative 7. The constant added outside is k, which is 2. Substituting negative 7 gives an output of 2, confirming the vertex.
Step zero
Discussion prompt
Worked example: build the equation from a vertex and one more point — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Start in vertex form and drop the vertex in
Answer:
Worked example
A parabola has its vertex at the point two across and three down, and it also passes through the point four across and five up. Find its rule.
\[ \text{vertex } (2,\ -3), \qquad \text{passes through } (4,\ 5) \]
Start in vertex form and drop the vertex in
Why: Vertex form is built out of the vertex, so two of the three unknowns are already handed to you. Only the stretch factor is still missing.
\[ f(x) = a(x - 2)^2 - 3 \]
Use the extra point to pin down the stretch factor
Why: A point on the graph means that input really does produce that output, so substituting both coordinates gives one equation in one unknown.
\[ 5 = a(4 - 2)^2 - 3 \]
Simplify and solve for the stretch factor
Why: Square first by order of operations, then undo the subtraction and the multiplication in turn.
\[ 5 = 4a - 3 \;\Rightarrow\; 8 = 4a \;\Rightarrow\; a = 2 \]
Write the finished model, in both forms
Why: Vertex form answers graphing questions; the expanded standard form is what most homework asks you to report.
\[ f(x) = 2(x - 2)^2 - 3 = 2x^2 - 8x + 5 \]
Verify both given facts in the finished rule
Why: The vertex input must return the vertex output, and the extra point must land on the curve. Both check out, so the model is right.
\[ f(2) = 2(0)^2 - 3 = -3, \qquad f(4) = 2(2)^2 - 3 = 8 - 3 = 5 \]
Picture it
Animation
Shows: Each line of the worked example "build the equation from a vertex and one more point", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The vertex input must return the vertex output, and the extra point must land on the curve. Both check out, so the model is right.
Section
Part 3
Concept
Almost every quadratic you meet in the wild arrives in standard form, not vertex form.
\[ f(x) = ax^2 + bx + c \]
This form is generous with some information and silent about the rest.
| Question | Standard form | Vertex form |
|---|---|---|
| Which way does it open? | yes - the sign of a | yes - the sign of a |
| Where is the vertical intercept? | yes - it is c | no, you must substitute |
| Where is the vertex? | no | yes, immediately |
| Where does it cross the axis? | solve, or factor | solve, one square root away |
So the whole job of this part of the deck is to get the vertex out of standard form. There are two routes: a formula, and completing the square. Start with the formula.
Comparison
Comparison matrix
From Standard form hides the vertex: refill the Vertex form column from what you know. The rest of the table is as it appeared.
| Question | Standard form | Vertex form |
|---|---|---|
| Which way does it open? | yes - the sign of a | yes - the sign of a |
| Where is the vertical intercept? | yes - it is c | no, you must substitute |
| Where is the vertex? | no | yes, immediately |
| Where does it cross the axis? | solve, or factor | solve, one square root away |
Intuition
The parabola is symmetric, so if it crosses the horizontal axis twice, those two crossings are mirror images. The mirror line has to run down the middle of them.
\[ f(x) = x^2 - 6x + 8 = (x - 2)(x - 4) \]
This one crosses at two places. Their midpoint is the average of the two.
\[ \frac{2 + 4}{2} = 3 \]
Now look at where that three came from in the original coefficients. The quadratic formula puts both roots the same distance either side of one central number, and that central number is what survives the averaging.
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \;\longrightarrow\; \text{average} = \frac{-b}{2a} \]
The plus-or-minus part cancels in the average. Whatever is left is the axis of symmetry - and it works even when there are no real crossings at all, because the center of symmetry does not care.
Concept
That average is worth memorizing on its own. It is the single most useful formula in this deck.
\[ x = \frac{-b}{2a} \]
vertex formula — The input at which a quadratic in standard form reaches its vertex: the opposite of the middle coefficient, divided by twice the leading coefficient.
The same number is the axis of symmetry, written as an equation of a vertical line.
\[ x = \frac{-b}{2a} \qquad \text{is the axis of symmetry} \]
Note what the formula does not give you: the second coordinate. The formula produces an input only.
Sorting
Sort into buckets
These are the pieces of Quadratic Functions, Parabolas, and Optimization, out of order. Put each one back under the part of the lesson it belongs to.
Picture it
Animation
Shows: Where the vertex is — a rendered Manim animation.
Rendered with Manim.
Takeaway: It comes straight out of completing the square.
Concept
Once you have the input, you get the output the ordinary way: feed it back into the function.
\[ \text{vertex} = \left( \frac{-b}{2a}, \ f\!\left( \frac{-b}{2a} \right) \right) \]
That nested notation looks heavy, but the action is simple: compute the number, then substitute the number.
Estimation
Predict first
Find the vertex, the axis of symmetry, and the minimum value.
Commit before you compute: what does Worked example: vertex and axis from standard form come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify with the symmetry test
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Inputs one unit either side of a true axis of symmetry must give equal outputs - and both must sit above the minimum for an upward parabola.
Worked example
Find the vertex, the axis of symmetry, and the minimum value.
\[ f(x) = 2x^2 - 12x + 7 \]
Name the three coefficients before touching the formula
Why: Most vertex-formula errors are bookkeeping errors. Writing the coefficients down with their signs prevents them.
\[ a = 2, \qquad b = -12, \qquad c = 7 \]
Substitute into the vertex formula
Why: The formula wants the opposite of the middle coefficient on top. The middle coefficient is negative twelve, so its opposite is positive twelve.
\[ x = \frac{-(-12)}{2(2)} = \frac{12}{4} = 3 \]
Substitute three back into the original function
Why: The formula gave the input; the output has to be computed. Square first, then multiply, then add.
\[ f(3) = 2(3)^2 - 12(3) + 7 = 18 - 36 + 7 = -11 \]
State all three answers in the right shapes
Why: The vertex is a point, the axis is an equation, and the minimum value is a single number. Homework grades all three differently.
\[ \text{vertex } (3,\ -11), \qquad x = 3, \qquad \text{minimum value } -11 \]
Verify with the symmetry test
Why: Inputs one unit either side of a true axis of symmetry must give equal outputs - and both must sit above the minimum for an upward parabola. They do.
\[ f(2) = 8 - 24 + 7 = -9, \qquad f(4) = 32 - 48 + 7 = -9 \]
Picture it
Animation
Shows: Each line of the worked example "vertex and axis from standard form", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Inputs one unit either side of a true axis of symmetry must give equal outputs - and both must sit above the minimum for an upward parabola. They do.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Copying the middle coefficient into the formula with the sign it already has.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: It is a positive six sitting right there, so it feels natural to divide it by twice the leading coefficient.
The formula asks for the opposite of the middle coefficient. Write the minus sign down before you substitute.
Why: It is a positive six sitting right there, so it feels natural to divide it by twice the leading coefficient.
Trap
Copying the middle coefficient into the formula with the sign it already has.
\[ f(x) = x^2 + 6x + 1 \]
Use the middle coefficient as written
Why: It is a positive six sitting right there, so it feels natural to divide it by twice the leading coefficient.
\[ x = \frac{6}{2(1)} = 3 \quad \Rightarrow \quad f(3) = 9 + 18 + 1 = 28 \]
The symmetry test refuses it
Why: Inputs on either side of a real axis of symmetry give equal outputs. These do not - and worse, both are below twenty-eight, which is impossible for a minimum on an upward parabola.
\[ f(2) = 17, \qquad f(4) = 41 \]
The formula asks for the opposite of the middle coefficient. Write the minus sign down before you substitute.
\[ x = \frac{-b}{2a} \]
Substitute the opposite of positive six
Why: The opposite of six is negative six, so the input is negative three. A positive middle coefficient pushes the vertex to the left.
\[ x = \frac{-6}{2(1)} = -3 \quad \Rightarrow \quad f(-3) = 9 - 18 + 1 = -8 \]
Verify with the symmetry test
Why: Now the two neighbours match exactly and both sit above the vertex output, which is what a minimum has to look like.
\[ f(-4) = -7, \qquad f(-2) = -7 \quad \Rightarrow \quad \text{vertex } (-3,\ -8) \]
Translation
\( f(-4) = -7, \qquad f(-2) = -7 \quad \Rightarrow \quad \text{vertex } (-3,\ -8) \)
Draw it
Translate both ways. First write the expression above as a sentence with no symbols in it at all. Then cover it, and write your sentence back as notation. If the two versions disagree, the disagreement is the thing to fix.
Missing information
Discussion prompt
Give the direction, the vertex, the axis of symmetry, the maximum value, and both kinds of intercept.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
There is an invisible negative one in front of the squared term. Negative means the parabola opens down, so its vertex will be a maximum.
Worked example
Give the direction, the vertex, the axis of symmetry, the maximum value, and both kinds of intercept.
\[ f(x) = -x^2 + 6x - 5 \]
Read the direction off the leading coefficient
Why: There is an invisible negative one in front of the squared term. Negative means the parabola opens down, so its vertex will be a maximum.
\[ a = -1, \qquad b = 6, \qquad c = -5 \]
Apply the vertex formula, watching both minus signs
Why: The opposite of six is negative six on top; twice negative one is negative two on the bottom. A negative divided by a negative is positive.
\[ x = \frac{-6}{2(-1)} = \frac{-6}{-2} = 3 \]
Substitute to get the maximum value
Why: Square the three first, then apply the leading negative. Skipping the order of operations here is the most common arithmetic slip.
\[ f(3) = -(3)^2 + 6(3) - 5 = -9 + 18 - 5 = 4 \]
Get the vertical intercept by substituting zero
Why: The constant term is exactly the output at zero, so this one is free in standard form.
\[ f(0) = -5 \]
Get the horizontal intercepts by setting the output to zero and factoring
Why: Multiplying through by negative one makes the leading coefficient positive and the trinomial easy to factor; the solutions are unchanged because zero times negative one is still zero.
\[ -x^2 + 6x - 5 = 0 \;\Rightarrow\; x^2 - 6x + 5 = 0 \;\Rightarrow\; (x-1)(x-5) = 0 \]
\[ x = 1 \quad \text{or} \quad x = 5 \]
Verify the crossings and the axis together
Why: Both candidates return zero in the original rule, and their midpoint is three - exactly the axis the vertex formula produced. Two independent facts agreeing is a real check.
\[ f(1) = -1 + 6 - 5 = 0, \quad f(5) = -25 + 30 - 5 = 0, \quad \frac{1+5}{2} = 3 \]
Picture it
Animation
Shows: Each line of the worked example "a full analysis of a downward parabola", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Both candidates return zero in the original rule, and their midpoint is three - exactly the axis the vertex formula produced. Two independent facts agreeing is a real check.
Constraint
Discussion prompt
Run Pattern: standard form to vertex, every time with this step confiscated:
Direction. Positive leading coefficient means a minimum at the vertex; negative means a maximum.
Is it still possible? If it is, say what takes its place and what it costs you. If it is not, say exactly what that step was providing that nothing else does.
Hint: A step you can drop for free was never load-bearing. If you cannot drop it, name the thing that goes wrong the moment it is gone.
Answer:
Pattern
\[ f(x) = ax^2 + bx + c \]
The symmetry check at the end catches every sign error the formula can produce, and it costs about fifteen seconds.
Check
Label the coefficients, run the formula, then substitute. Do it on paper before you pick.
\[ f(x) = 3x^2 + 18x + 4 \]
Check your understanding
What is the vertex of this parabola?
Answer: A
Why: The vertex formula gives the opposite of 18 over twice 3, which is negative 18 over 6, or negative 3. Substituting negative 3 gives 3 times 9, minus 54, plus 4, which is 27 minus 54 plus 4, or negative 23. So the vertex is at negative 3 and negative 23.
Concept
You can square any real number. There is no division and no even root in a quadratic, so nothing is ever forbidden as an input.
\[ \text{domain} = (-\infty,\ \infty) \]
That answer is the same for every quadratic function in this deck, with no work required. Applied problems are the one exception, and only because the story restricts the inputs, not the algebra.
Picture it
Animation
Shows: Why revenue is often a parabola — a rendered Manim animation.
Rendered with Manim.
Takeaway: Price times quantity, when raising the price loses customers.
Concept
The outputs are a different story. A parabola turns around, so the outputs stop at the vertex and never come back past it.
If it opens up, the vertex output is the smallest value the function ever produces, and everything above it is reachable.
\[ a > 0: \quad \text{range} = [\,k,\ \infty) \]
If it opens down, the vertex output is the largest value, and everything below it is reachable.
\[ a < 0: \quad \text{range} = (-\infty,\ k\,] \]
The bracket is square on the vertex side because the vertex output really is attained - the graph touches it. The infinite end always gets a round parenthesis.
Step zero
Discussion prompt
Worked example: domain and range from standard form — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Write the domain immediately
Answer:
Worked example
State the domain and the range in interval notation.
\[ f(x) = -3x^2 + 12x - 5 \]
Write the domain immediately
Why: It is a quadratic, so no input is excluded. This needs no computation at all.
\[ \text{domain} = (-\infty,\ \infty) \]
Decide the direction from the leading coefficient
Why: Negative three is negative, so the parabola opens down. The range will therefore run downward from the vertex output.
Find the vertex input with the formula
Why: The opposite of twelve is negative twelve; twice negative three is negative six. Two negatives divide to a positive.
\[ x = \frac{-12}{2(-3)} = \frac{-12}{-6} = 2 \]
Substitute to get the largest output
Why: Square the two first, then multiply by negative three, then finish left to right.
\[ f(2) = -3(2)^2 + 12(2) - 5 = -12 + 24 - 5 = 7 \]
Write the range as an interval closed on the vertex side
Why: Seven is actually reached, so the bracket is square; the outputs run down forever from there, so the other end is an infinite parenthesis.
\[ \text{range} = (-\infty,\ 7\,] \]
Verify that no input beats seven
Why: Two inputs on either side of the axis give equal outputs, and both are below seven - exactly what a maximum requires.
\[ f(1) = -3 + 12 - 5 = 4, \qquad f(3) = -27 + 36 - 5 = 4 \]
Picture it
Animation
Shows: Each line of the worked example "domain and range from standard form", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Two inputs on either side of the axis give equal outputs, and both are below seven - exactly what a maximum requires.
Concept
The vertex is a point, so it carries two numbers, and applied problems almost always want only one of them.
| The question asks | The answer is | Which coordinate |
|---|---|---|
| At what price is revenue highest? | the input | first |
| What is the highest revenue? | the output | second |
| When does the ball peak? | the input | first |
| How high does the ball get? | the output | second |
Underline the question word before you compute anything. Where and when ask for the first coordinate; what value, how high, and how much ask for the second.
maximum value — The largest output the function ever produces - the second coordinate of the vertex. The input that achieves it is a separate answer, sometimes called the maximizer.
Matching
Match the pairs
Match each term to the definition this lesson gave it — not the one you would guess from the word.
Why: These are the working definitions of parabola, vertex, axis of symmetry, vertex formula, maximum value as Quadratic Functions, Parabolas, and Optimization uses them. Pairing them correctly is the test of whether you could state each one with the slide switched off.
Anomaly
Predict first
A student writes this, and it looks reasonable:
A shop models its daily revenue in dollars against the price it charges in dollars. What is the maximum revenue?
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The formula produced a number, the work felt finished, and twenty is a perfectly reasonable-looking answer.
The vertex formula gives the price. Revenue is what the function returns, so there is one more substitution to do.
Why: The formula produced a number, the work felt finished, and twenty is a perfectly reasonable-looking answer.
Trap
A shop models its daily revenue in dollars against the price it charges in dollars. What is the maximum revenue?
\[ R(p) = -5p^2 + 200p \]
Run the vertex formula and stop
Why: The formula produced a number, the work felt finished, and twenty is a perfectly reasonable-looking answer.
\[ p = \frac{-200}{2(-5)} = 20 \]
Report twenty dollars as the maximum revenue
Why: But twenty is a price, not a revenue. It answers a question nobody asked, and it is off by a factor of a hundred.
The vertex formula gives the price. Revenue is what the function returns, so there is one more substitution to do.
\[ R(p) = -5p^2 + 200p \]
Use the formula output as an input
Why: Twenty is the price that maximizes revenue. To learn the revenue itself, feed it to the function.
\[ R(20) = -5(20)^2 + 200(20) = -2000 + 4000 = 2000 \]
Verify by testing prices on either side
Why: Both neighbouring prices produce less revenue than two thousand dollars, so twenty really is the peak - and the answer to the question asked is the two thousand, at a price of twenty.
\[ R(19) = 1995, \qquad R(21) = 1995 \]
Break the constraint
Discussion prompt
The rule this trap just fixed:
Twenty is the price that maximizes revenue. To learn the revenue itself, feed it to the function.
Now break it on purpose. Build a case that violates it and follow the consequences until something visibly fails. Where does the failure first show up — and would you have noticed it if you had not been looking?
Hint: The dangerous rules are the ones whose violation still produces an answer. If yours fails loudly, try to find one that fails quietly.
Answer:
The formula produced a number, the work felt finished, and twenty is a perfectly reasonable-looking answer.
Check
Find the vertex first, then decide which way the outputs run.
\[ f(x) = 2x^2 - 8x + 3 \]
Check your understanding
What is the range of this function?
Answer: A
Why: The vertex input is the opposite of negative 8 over twice 2, which is 8 over 4, or 2. Substituting gives 8 minus 16 plus 3, which is negative 5. The leading coefficient 2 is positive, so the parabola opens up and negative 5 is the smallest output.
Section
Part 4
Picture it
Figure (svg): A large square with two shaded rectangular strips attached along its right side and its bottom, leaving an empty dashed square in the bottom-right corner that would complete the whole shape.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
The name is not a metaphor. Picture the expression as area: a square whose side is the variable, plus two identical strips glued to two of its sides.
Intuition
The name is not a metaphor. Picture the expression as area: a square whose side is the variable, plus two identical strips glued to two of its sides.
Figure (svg): A large square with two shaded rectangular strips attached along its right side and its bottom, leaving an empty dashed square in the bottom-right corner that would complete the whole shape.
The middle term is the two strips. Split it evenly, one strip per side, and each strip has the same width: half the middle coefficient.
Now the picture is a square with a bite out of the corner. The bite is a small square whose side is that same half - so its area is the half, squared. Add exactly that much and the shape closes up.
That is the whole method. Everything else is bookkeeping to make sure you give back whatever you borrowed.
Concept
Reading the picture backwards gives the algebraic pattern every perfect-square trinomial follows.
\[ x^2 + bx + \left( \frac{b}{2} \right)^{2} = \left( x + \frac{b}{2} \right)^{2} \]
So the number you need is never a mystery. Take the middle coefficient, halve it, square the result.
| Middle term | Half of it | Squared | Completed square |
|---|---|---|---|
| 8x | 4 | 16 | (x + 4)^2 |
| -6x | -3 | 9 | (x - 3)^2 |
| 10x | 5 | 25 | (x + 5)^2 |
| -x | -1/2 | 1/4 | (x - 1/2)^2 |
Notice the sign in the finished square is the sign of the half, not of the squared number. The squared number is always positive.
But you cannot simply add a number to a function and pretend nothing happened. Whatever you add, you must subtract right back.
Pattern
Step through it
Step through Half the middle coefficient, then square it one row at a time. What is driving the change, and what would the row after the last one be?
Estimation
Predict first
Rewrite in vertex form and state the vertex.
Commit before you compute: what does Worked example: completing the square, leading coefficient… come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify by expanding back and by the vertex formula
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Expanding returns the original expression exactly, and the vertex formula gives the same input independently.
Worked example
Rewrite in vertex form and state the vertex.
\[ f(x) = x^2 + 8x + 11 \]
Set the constant aside and leave a gap after the middle term
Why: Only the squared and middle terms take part in completing the square. The constant is a spectator until the very end.
\[ f(x) = \left( x^2 + 8x + \underline{\phantom{16}} \right) + 11 \]
Halve the middle coefficient and square it
Why: Half of eight is four, and four squared is sixteen. That sixteen is the corner the picture is missing.
\[ \left( \frac{8}{2} \right)^{2} = 4^2 = 16 \]
Add sixteen and subtract sixteen in the same line
Why: Adding and subtracting the same amount changes nothing about the function - it only regroups it. This is the step that keeps the equation honest.
\[ f(x) = \left( x^2 + 8x + 16 \right) - 16 + 11 \]
Collapse the bracket into a square and combine the loose constants
Why: The bracket is now a perfect-square trinomial whose square root is the variable plus the half. Outside, negative sixteen plus eleven is negative five.
\[ f(x) = (x + 4)^2 - 5 \]
Read the vertex, remembering the built-in minus
Why: The plus four inside is really minus a negative four, so the horizontal coordinate is negative four; the constant outside is the vertical one.
\[ \text{vertex } (-4,\ -5) \]
Verify by expanding back and by the vertex formula
Why: Expanding returns the original expression exactly, and the vertex formula gives the same input independently. Two agreeing checks means the conversion is right.
\[ (x+4)^2 - 5 = x^2 + 8x + 16 - 5 = x^2 + 8x + 11, \qquad \frac{-8}{2(1)} = -4 \]
Picture it
Animation
Shows: Height under gravity — a rendered Manim animation.
Rendered with Manim.
Takeaway: The vertex is the highest point; the zeros are launch and landing.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Adding the completing number and moving on, because the bracket now looks right.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The bracket really is a perfect square now, so it feels like progress - but sixteen has been created out of nothing.
Anything you add must be subtracted in the same line. You are regrouping the expression, not changing it.
Why: The bracket really is a perfect square now, so it feels like progress - but sixteen has been created out of nothing.
Trap
Adding the completing number and moving on, because the bracket now looks right.
\[ f(x) = x^2 + 8x + 11 \]
Add sixteen inside and keep the eleven outside
Why: The bracket really is a perfect square now, so it feels like progress - but sixteen has been created out of nothing.
\[ f(x) \stackrel{?}{=} (x + 4)^2 + 11 \]
Test it at zero: the two rules disagree
Why: The original returns eleven at an input of zero; the rewritten version returns twenty-seven. They are not the same function, so the rewrite is wrong.
\[ \text{original: } f(0) = 11 \qquad \text{rewrite: } (0+4)^2 + 11 = 27 \]
Anything you add must be subtracted in the same line. You are regrouping the expression, not changing it.
\[ f(x) = x^2 + 8x + 11 \]
Add sixteen and subtract sixteen together
Why: The net change is zero, so the function is untouched - but the first three terms can now be folded into a square.
\[ f(x) = \left( x^2 + 8x + 16 \right) - 16 + 11 = (x+4)^2 - 5 \]
Verify at zero: the two rules now agree
Why: Both return eleven at an input of zero, which is what an honest rewrite must do at every input.
\[ (0+4)^2 - 5 = 16 - 5 = 11 \]
Notation
Annotate
From Trap: adding the corner without paying it back — read this one piece at a time. What is each part doing?
On: \( (0+4)^2 - 5 = 16 - 5 = 11 \)
Concept
The half-then-square pattern only works when the squared term stands alone. So the first move is to pull the leading coefficient out of the first two terms.
\[ ax^2 + bx + c = a\left( x^2 + \frac{b}{a}x \right) + c \]
The constant stays outside the parentheses. It is not part of the square, so dragging it in only creates fractions you do not need.
Then comes the step everyone forgets. The number you add sits inside the parentheses, so it gets multiplied by the leading coefficient on its way out. You must give back that multiplied amount, not the bare number.
\[ a\left( x^2 + \tfrac{b}{a}x + m \right) = a\left( x^2 + \tfrac{b}{a}x \right) + am \]
So the compensation is the leading coefficient times the number you added. Write that product out explicitly rather than doing it in your head.
Hypothesis
Predict first
Worked example: completing the square with a positive stretch is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.
Correct: Factor the three out of the first two terms only
Why: Twelve divided by three is four, and the sign comes along. The constant five stays outside, untouched.
A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.
Worked example
Rewrite in vertex form and state the minimum value.
\[ f(x) = 3x^2 - 12x + 5 \]
Factor the three out of the first two terms only
Why: Twelve divided by three is four, and the sign comes along. The constant five stays outside, untouched.
\[ f(x) = 3\left( x^2 - 4x \right) + 5 \]
Halve the new middle coefficient and square it
Why: Inside the parentheses the middle coefficient is negative four. Half of that is negative two, and negative two squared is positive four.
\[ \left( \frac{-4}{2} \right)^{2} = (-2)^2 = 4 \]
Add four inside and subtract three times four outside
Why: The four is inside a bracket multiplied by three, so it really adds twelve to the function. Twelve is what must be given back.
\[ f(x) = 3\left( x^2 - 4x + 4 \right) - 3(4) + 5 \]
Collapse the square and combine the constants
Why: The bracket folds to the variable minus two, squared, because negative two was the half. Outside, negative twelve plus five is negative seven.
\[ f(x) = 3(x - 2)^2 - 7 \]
Read the vertex and the minimum value
Why: The minus is already showing, so the coordinates are two and negative seven. The leading three is positive, so that output is a minimum.
\[ \text{vertex } (2,\ -7), \qquad \text{minimum value } -7 \]
Verify by expanding back to the original
Why: Distributing the three and combining constants must reproduce the starting expression term for term - and it does.
\[ 3(x-2)^2 - 7 = 3x^2 - 12x + 12 - 7 = 3x^2 - 12x + 5 \]
Picture it
Animation
Shows: Each line of the worked example "completing the square with a positive stretch", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Distributing the three and combining constants must reproduce the starting expression term for term - and it does.
Ranking
Put in order
Put the moves of Worked example: completing the square with a negative stretch into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Twelve divided by negative two is negative six, so the middle sign flips inside the parentheses.
Worked example
Rewrite in vertex form and state the maximum value. Watch every sign.
\[ f(x) = -2x^2 + 12x - 13 \]
Factor negative two out of the first two terms
Why: Twelve divided by negative two is negative six, so the middle sign flips inside the parentheses. This flip is where most errors start.
\[ f(x) = -2\left( x^2 - 6x \right) - 13 \]
Halve the inside middle coefficient and square it
Why: Half of negative six is negative three, and negative three squared is positive nine.
\[ \left( \frac{-6}{2} \right)^{2} = (-3)^2 = 9 \]
Add nine inside and give back negative two times nine
Why: Adding nine inside a bracket multiplied by negative two actually subtracts eighteen from the function, so eighteen must be added back outside.
\[ f(x) = -2\left( x^2 - 6x + 9 \right) + 18 - 13 \]
Collapse and combine
Why: The bracket folds to the variable minus three, squared; outside, eighteen minus thirteen is five.
\[ f(x) = -2(x - 3)^2 + 5 \]
Read the vertex and the maximum value
Why: The stretch factor is negative, so the parabola opens down and the vertex output is the largest value the function reaches.
\[ \text{vertex } (3,\ 5), \qquad \text{maximum value } 5 \]
Verify by expanding and by substituting three into the original
Why: The expansion reproduces the original, and the original evaluated at three really does give five. The conversion survives both tests.
\[ -2(x-3)^2 + 5 = -2x^2 + 12x - 18 + 5 = -2x^2 + 12x - 13, \qquad f(3) = -18 + 36 - 13 = 5 \]
Pattern
\[ ax^2 + bx + c \;\longrightarrow\; a(x-h)^2 + k \]
Then verify, always, by expanding your vertex form back out. If it does not reproduce the original expression, a sign slipped somewhere in step three.
Picture it
Animation
Shows: Standard to vertex form — a rendered Manim animation.
Rendered with Manim.
Takeaway: Complete the square, then compensate for what you added.
Elimination
Eliminate the wrong options
Which is the correct vertex form of this function?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: Factoring gives 2 times the quantity x squared plus 6x, plus 7. Half of 6 is 3 and 3 squared is 9, and that 9 sits inside a bracket multiplied by 2, so 18 must be subtracted. Then 7 minus 18 is negative 11. Expanding 2 times the square of x plus 3, minus 11 gives 2x squared plus 12x plus 18 minus 11, which is the original.
Check
Factor first, then halve and square, then remember what you owe back.
\[ f(x) = 2x^2 + 12x + 7 \]
Check your understanding
Which is the correct vertex form of this function?
Answer: A
Why: Factoring gives 2 times the quantity x squared plus 6x, plus 7. Half of 6 is 3 and 3 squared is 9, and that 9 sits inside a bracket multiplied by 2, so 18 must be subtracted. Then 7 minus 18 is negative 11. Expanding 2 times the square of x plus 3, minus 11 gives 2x squared plus 12x plus 18 minus 11, which is the original.
Section
Part 5
Concept
A graph meets the vertical axis where the input is zero. Substituting zero into standard form wipes out both terms that carry the variable.
\[ f(0) = a(0)^2 + b(0) + c = c \]
So the constant term is the vertical intercept. No work, no risk of error, and it is a useful sanity point when you sketch.
In vertex form the same idea works, but you do have to substitute, because the input zero is not the special input any more.
\[ f(x) = a(x-h)^2 + k \;\Rightarrow\; f(0) = ah^2 + k \]
Concept
A graph meets the horizontal axis where the output is zero. Every question below is the same question wearing a different coat.
| The wording | What it asks for |
|---|---|
| Find the x-intercepts of the graph | the inputs where the output is zero |
| Find the zeros of the function | the inputs where the output is zero |
| Solve the equation for the variable | the inputs where the output is zero |
| Find the roots | the inputs where the output is zero |
\[ ax^2 + bx + c = 0 \]
All four wordings hand you this one equation. You already know three ways to solve it: factoring, the square root property, and the quadratic formula.
One difference worth naming: an intercept is a point, so it is written as a pair. A zero or a root is just the input number.
Comparison
Comparison matrix
From Horizontal intercepts, zeros, and roots are one idea with…: refill the What it asks for column from what you know. The rest of the table is as it appeared.
| The wording | What it asks for |
|---|---|
| Find the x-intercepts of the graph | the inputs where the output is zero |
| Find the zeros of the function | the inputs where the output is zero |
| Solve the equation for the variable | the inputs where the output is zero |
| Find the roots | the inputs where the output is zero |
Fill the middle
Fill in the blanks
From Worked example: all the intercepts by factoring — finish the line. Write what belongs on the right of the equals sign before you look.
f(x) = x^2 - 2x - 15
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. The output at an input of zero is the constant, so the graph crosses the vertical axis fifteen units below the origin.
Worked example
Find every intercept and the vertex.
\[ f(x) = x^2 - 2x - 15 \]
Write the vertical intercept straight off the constant term
Why: The output at an input of zero is the constant, so the graph crosses the vertical axis fifteen units below the origin.
\[ (0,\ -15) \]
Set the output to zero and factor the trinomial
Why: Look for two numbers that multiply to negative fifteen and add to negative two. Negative five and positive three do it.
\[ x^2 - 2x - 15 = 0 \;\Rightarrow\; (x - 5)(x + 3) = 0 \]
Apply the zero-product property
Why: A product is zero only when one of its factors is zero, so each factor gives one solution.
\[ x = 5 \quad \text{or} \quad x = -3 \;\Rightarrow\; (5,\ 0) \text{ and } (-3,\ 0) \]
Find the vertex with the formula
Why: The opposite of negative two over twice one is one. Substituting one gives one minus two minus fifteen.
\[ x = \frac{2}{2} = 1, \qquad f(1) = 1 - 2 - 15 = -16 \]
Verify the zeros and the axis against each other
Why: Both candidates return zero in the original rule, and their midpoint is one - the same axis the vertex formula produced. The two crossings are each four units from it, as symmetry demands.
\[ f(5) = 25 - 10 - 15 = 0, \quad f(-3) = 9 + 6 - 15 = 0, \quad \frac{5 + (-3)}{2} = 1 \]
Picture it
Animation
Shows: Each line of the worked example "all the intercepts by factoring", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Both candidates return zero in the original rule, and their midpoint is one - the same axis the vertex formula produced. The two crossings are each four units from it, as symmetry demands.
Step zero
Discussion prompt
Worked example: intercepts that do not factor — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Label the coefficients and set the output to zero
Answer:
Worked example
Find the horizontal intercepts. Nothing here factors over the integers, so reach for the formula.
\[ f(x) = x^2 - 6x + 4 \]
Label the coefficients and set the output to zero
Why: The quadratic formula only applies to an equation set equal to zero, and it needs all three coefficients with their signs.
\[ a = 1, \ b = -6, \ c = 4, \qquad x^2 - 6x + 4 = 0 \]
Substitute into the quadratic formula
Why: Put every substituted coefficient in parentheses. The opposite of negative six is positive six on top, and negative six squared is positive thirty-six.
\[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(4)}}{2(1)} = \frac{6 \pm \sqrt{36 - 16}}{2} \]
Simplify the radical before dividing
Why: Twenty has a factor of four, a perfect square, so two comes out of the root. Simplifying first is what makes the fraction reduce cleanly.
\[ x = \frac{6 \pm \sqrt{20}}{2} = \frac{6 \pm 2\sqrt{5}}{2} \]
Divide every term by two
Why: Both terms on top share a factor of two. Cancelling from only one of them is the classic error here.
\[ x = 3 \pm \sqrt{5} \]
Verify against the vertex and the constant term
Why: The two roots average to three, which is exactly the vertex formula answer, and their product is nine minus five, which equals the constant term four. Both facts confirm the pair.
\[ \frac{(3+\sqrt5) + (3-\sqrt5)}{2} = 3, \qquad (3+\sqrt5)(3-\sqrt5) = 9 - 5 = 4 \]
Concept
Inside the quadratic formula sits one expression that decides everything about the number of crossings: the part under the radical.
\[ D = b^2 - 4ac \]
discriminant — The quantity under the radical in the quadratic formula. Its sign alone tells you how many real solutions the equation has, and therefore how many times the parabola meets the horizontal axis.
It is cheap to compute - one squaring and one multiplication - and it can save you from hunting for crossings that are not there.
Concept
Figure (svg): Three small parabolas above a horizontal axis: the first dips below the axis and crosses it twice, the second just touches the axis at its lowest point, and the third sits entirely above the axis without meeting it.
A positive discriminant means the radical is a real nonzero number, so the plus-or-minus produces two different inputs.
\[ b^2 - 4ac > 0 \;\Longrightarrow\; \text{two real zeros, two crossings} \]
A discriminant of zero makes the radical vanish, so the plus and the minus give the same answer. The graph touches the axis exactly at its vertex.
\[ b^2 - 4ac = 0 \;\Longrightarrow\; \text{one repeated real zero, the vertex sits on the axis} \]
A negative discriminant asks for the square root of a negative number, which is not real. The parabola never reaches the axis at all.
\[ b^2 - 4ac < 0 \;\Longrightarrow\; \text{no real zeros, no crossings} \]
Explain it
Discussion prompt
Explain Three cases, three pictures to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
A positive discriminant means the radical is a real nonzero number, so the plus-or-minus produces two different inputs.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Reaching for the formula on autopilot and forcing an answer out of it.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Four minus twenty is negative sixteen, but sixteen is a familiar perfect square, so it is tempting to write its root as four and move on.
Compute the discriminant first. Its sign decides whether there is anything to find.
Why: Four minus twenty is negative sixteen, but sixteen is a familiar perfect square, so it is tempting to write its root as four and move on.
Trap
Reaching for the formula on autopilot and forcing an answer out of it.
\[ f(x) = x^2 + 2x + 5 \]
Compute under the radical and keep going anyway
Why: Four minus twenty is negative sixteen, but sixteen is a familiar perfect square, so it is tempting to write its root as four and move on.
\[ x = \frac{-2 \pm \sqrt{4 - 20}}{2} = \frac{-2 \pm \sqrt{-16}}{2} \stackrel{?}{=} \frac{-2 \pm 4}{2} \]
Report crossings at one and negative three
Why: Substituting either candidate exposes the fiction: neither returns zero, because the negative under the radical was quietly discarded.
\[ f(1) = 8, \qquad f(-3) = 8 \]
Compute the discriminant first. Its sign decides whether there is anything to find.
\[ b^2 - 4ac = (2)^2 - 4(1)(5) = 4 - 20 = -16 \]
Negative discriminant means no horizontal intercepts
Why: There is no real square root of a negative number, so the plus-or-minus never produces a real input. The graph misses the axis entirely.
Verify with the vertex
Why: The vertex input is negative one and the output there is four. The parabola opens up, so four is the smallest output it ever has - it can never get down to zero.
\[ x = \frac{-2}{2} = -1, \qquad f(-1) = 1 - 2 + 5 = 4 > 0 \]
Two truths and a lie
Sort into buckets
Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.
Fill the middle
Fill in the blanks
From Worked example: count first, then prove it with the vertex — finish the line. Write what belongs on the right of the equals sign before you look.
f(x) = 2x^2 - 4x + 5
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Negative four squared is positive sixteen; four times two times five is forty.
Worked example
How many times does this graph meet the horizontal axis?
\[ f(x) = 2x^2 - 4x + 5 \]
Compute the discriminant
Why: Negative four squared is positive sixteen; four times two times five is forty. Sixteen minus forty is negative twenty-four.
\[ b^2 - 4ac = (-4)^2 - 4(2)(5) = 16 - 40 = -24 \]
Read the count off the sign
Why: The discriminant is negative, so there are no real zeros and the graph never meets the horizontal axis.
\[ -24 < 0 \;\Rightarrow\; \text{zero crossings} \]
Convert to vertex form for a second opinion
Why: Factor the two out of the first two terms, halve negative two to get negative one, square it to get one, and give back two times one.
\[ f(x) = 2(x^2 - 2x + 1) - 2 + 5 = 2(x-1)^2 + 3 \]
Read the minimum value
Why: The squared term is never negative and the stretch factor is positive, so the smallest the whole expression can be is three, reached at an input of one.
\[ \text{minimum value } 3 \]
Verify that the two answers agree
Why: A function whose smallest output is three can never output zero, which is exactly what a negative discriminant claimed. Expanding the vertex form also returns the original expression.
\[ 2(x-1)^2 + 3 = 2x^2 - 4x + 2 + 3 = 2x^2 - 4x + 5 \]
Picture it
Animation
Shows: Each line of the worked example "count first, then prove it with the vertex", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: A function whose smallest output is three can never output zero, which is exactly what a negative discriminant claimed. Expanding the vertex form also returns the original expression.
Ranking
Put in order
These are the steps of Pattern: sketch any parabola end to end, scrambled. Put them back in order before the next slide shows you.
Why: This is the order the recipe itself gives. Recalling the sequence without the slide in front of you is the difference between recognising the method and being able to run it — most of what goes wrong in practice is a step done out of turn.
Pattern
Six pieces of information are enough to draw a convincing graph by hand. Collect them in this order.
Then draw a smooth curve through the plotted points. The mirror trick in step four means five marks cost you only three computations.
Finally, sanity-check the picture: the vertex must be the lowest or highest point drawn, and the two arms must be the same height at equal distances from the axis.
Edge cases
Discussion prompt
Pattern: sketch any parabola end to end works on the cases you have just seen. Push it to the edge: what is the most degenerate input it still handles — empty, zero, one item, everything equal — and what is the first case where it stops being true? Name the case, not just "it breaks".
Hint: Try the smallest legal input, then the largest, then the one where two things collide. Methods are specified at their edges; the middle takes care of itself.
Answer:
Six pieces of information are enough to draw a convincing graph by hand. Collect them in this order.
Prediction
Predict first
How many times does the graph of this function meet the horizontal axis?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: exactly once
Why: The discriminant is negative 10 squared minus 4 times 1 times 25, which is 100 minus 100, or 0. A discriminant of zero means the plus and the minus branches give the same input, so there is one repeated zero. Here the function is the square of x minus 5, whose vertex sits at 5 and 0, right on the axis.
Check
Compute the discriminant before you try to solve anything.
\[ f(x) = x^2 - 10x + 25 \]
Check your understanding
How many times does the graph of this function meet the horizontal axis?
Answer: A
Why: The discriminant is negative 10 squared minus 4 times 1 times 25, which is 100 minus 100, or 0. A discriminant of zero means the plus and the minus branches give the same input, so there is one repeated zero. Here the function is the square of x minus 5, whose vertex sits at 5 and 0, right on the axis.
Concept
You now have both forms and a way to travel between them. Choose based on what the question wants.
| You want | Best form | Move |
|---|---|---|
| the vertex | vertex form | read it, flipping the inside sign |
| the vertex, from standard form | standard form | the vertex formula, then substitute |
| the vertical intercept | standard form | it is the constant term |
| the horizontal intercepts | either | factor, formula, or square root property |
| the number of crossings | standard form | the discriminant |
| a model from a vertex and a point | vertex form | substitute the point, solve for the stretch |
Completing the square is the bridge in one direction; expanding is the bridge back. Neither changes the parabola - only the packaging.
Discrimination
Sort into buckets
Sort these by Best form, from memory, without looking back at The two forms, side by side. Telling them apart on the spot is the skill; the table is only where the answer happens to be written down.
Section
Part 6
Intuition
Businesses, engineers, and farmers all ask the same shaped question: I have a fixed amount of something, how do I get the most out of it?
The reason quadratics show up so often is a tug-of-war. Raise the price and each sale earns more, but you make fewer sales. Make the garden longer and you gain along one direction, but the fence you spent on length is fence you cannot spend on width.
Two quantities pulling against each other, multiplied together, produce a squared term. The product climbs, peaks, and falls - which is exactly a parabola opening down.
So the peak of the trade-off is the vertex, and you already know four ways to find one. The hard part of these problems is never the algebra. It is writing the function.
Analogy
Discussion prompt
Explain Every optimization question is a vertex question by analogy to something with no College Algebra in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Businesses, engineers, and farmers all ask the same shaped question: I have a fixed amount of something, how do I get the most out of it?
Picture it
Animation
Shows: The area function has one peak — a rendered Manim animation.
Rendered with Manim.
Takeaway: Constraint substituted in, the problem becomes finding one vertex.
Concept
Applied quadratics feel unpredictable until you notice that the setup is always the same four moves.
Move two is where the fixed amount gets used. If a problem gives you a total and you never subtract from it, you have skipped the constraint and your function will not be quadratic.
Move four is where points are lost. Go back and reread the question before writing the final line.
Counterexample
Discussion prompt
Applied quadratics feel unpredictable until you notice that the setup is always the same four moves.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
Move two is where the fixed amount gets used. If a problem gives you a total and you never subtract from it, you have skipped the constraint and your function will not be quadratic.
Estimation
Predict first
You have one hundred twenty feet of fencing and a long straight barn wall. You want a rectangular garden using the barn as one full side, so you only fence the other three. What dimensions give the largest area, and how large is it?
Commit before you compute: what does Worked example: the largest garden a fence can hold come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify the constraint and test the neighbours
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. The three fenced sides really do add to one hundred twenty feet, and shifting the width one foot either way loses area.
Worked example
You have one hundred twenty feet of fencing and a long straight barn wall. You want a rectangular garden using the barn as one full side, so you only fence the other three. What dimensions give the largest area, and how large is it?
Name one variable with units
Why: The two sides perpendicular to the barn are equal, so calling that common length the variable describes two of the three fenced sides at once.
\[ w = \text{length of each side perpendicular to the barn, in feet} \]
Use the fencing total to express the remaining side
Why: Three sides are fenced: two of length w and one parallel to the barn. Their total must be exactly one hundred twenty feet, so the third side is whatever is left.
\[ 2w + \ell = 120 \;\Rightarrow\; \ell = 120 - 2w \]
Write the area as a function of that one variable
Why: Area is length times width. Substituting the expression for the remaining side turns a two-variable formula into a one-variable quadratic.
\[ A(w) = w(120 - 2w) = -2w^2 + 120w \]
Note the direction and find the vertex input
Why: The leading coefficient is negative, so this opens down and the vertex really is a maximum. The opposite of one hundred twenty over twice negative two is thirty.
\[ w = \frac{-120}{2(-2)} = \frac{-120}{-4} = 30 \]
Get the remaining side and the maximum area
Why: Thirty is a width, not an area. The remaining side comes from the constraint, and the area comes from substituting into the area function.
\[ \ell = 120 - 2(30) = 60, \qquad A(30) = 30(60) = 1800 \]
Verify the constraint and test the neighbours
Why: The three fenced sides really do add to one hundred twenty feet, and shifting the width one foot either way loses area. So thirty feet by sixty feet, with a maximum area of one thousand eight hundred square feet.
\[ 2(30) + 60 = 120, \qquad A(29) = 29(62) = 1798, \qquad A(31) = 31(58) = 1798 \]
Concept
The algebra will happily accept a garden width of negative five feet. The situation will not.
In the fencing problem, the width has to be positive, and the remaining side has to be positive too, which caps it.
\[ w > 0 \quad \text{and} \quad 120 - 2w > 0 \;\Longrightarrow\; 0 < w < 60 \]
The winning width of thirty feet sits comfortably inside that window, so the answer stands. Always check that the vertex input is actually allowed by the story.
Occasionally it is not - a factory that can produce at most eighty units, with a vertex at ninety. Then the best allowed value sits at the edge of the window, not at the vertex, and you evaluate there instead.
Picture it
Animation
Shows: Context restricts the domain — a rendered Manim animation.
Rendered with Manim.
Takeaway: The algebra allows values the situation forbids.
Step zero
Discussion prompt
Worked example: the price that maximizes revenue — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Name the variable as the number of one-dollar drops
Answer:
Worked example
A theater sells five hundred tickets a night at thirty dollars each. A survey says that for every one dollar the price drops, twenty-five more tickets sell. What price maximizes revenue, and what is that revenue?
Name the variable as the number of one-dollar drops
Why: Counting the drops rather than the price itself makes both the price and the ticket count easy to write, and keeps the arithmetic small.
\[ x = \text{number of one-dollar price reductions} \]
Write the price and the quantity in terms of that variable
Why: Each drop lowers the price by one and raises the count by twenty-five. This is the linear demand relationship the problem describes.
\[ \text{price} = 30 - x, \qquad \text{tickets} = 500 + 25x \]
Revenue is price times quantity - multiply and expand
Why: Multiplying two linear expressions produces the squared term. Expanding into standard form is what lets the vertex formula apply.
\[ R(x) = (30 - x)(500 + 25x) = 15000 + 750x - 500x - 25x^2 \]
\[ R(x) = -25x^2 + 250x + 15000 \]
Confirm it opens down before optimizing
Why: The leading coefficient is negative twenty-five, so the vertex is a maximum. If it had come out positive, the problem would have no largest revenue.
Find the vertex input
Why: The opposite of two hundred fifty over twice negative twenty-five is five, so five one-dollar reductions is the sweet spot.
\[ x = \frac{-250}{2(-25)} = \frac{-250}{-50} = 5 \]
Translate back into the language of the question
Why: Five is a count of price drops, not a price and not a revenue. Convert it into the price, the ticket count, and the revenue the question asked for.
\[ \text{price} = 25, \qquad \text{tickets} = 625, \qquad R(5) = 15625 \]
Verify by multiplying directly and by testing neighbours
Why: Twenty-five dollars times six hundred twenty-five tickets really is fifteen thousand six hundred twenty-five dollars, and one drop either side earns less. The answer is a price of twenty-five dollars for a revenue of fifteen thousand six hundred twenty-five dollars.
\[ 25 \cdot 625 = 15625, \qquad R(4) = 15600, \qquad R(6) = 15600 \]
Picture it
Animation
Shows: Each line of the worked example "the price that maximizes revenue", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Twenty-five dollars times six hundred twenty-five tickets really is fifteen thousand six hundred twenty-five dollars, and one drop either side earns less. The answer is a price of twenty-five dollars for a revenue of fifteen thousand six hundred twenty-five dollars.
Ranking
Put in order
Put the moves of Worked example: how high, and for how long into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The leading coefficient is negative, so the path peaks and comes back down.
Worked example
A ball is thrown upward from the top of an eighty-foot platform. Its height in feet after a number of seconds is modeled below. Find the greatest height it reaches, when it reaches it, and how long it stays in the air.
\[ h(t) = -16t^2 + 64t + 80 \]
Check the direction and the starting height
Why: The leading coefficient is negative, so the path peaks and comes back down. The constant term is the height at time zero, which matches the eighty-foot platform.
\[ h(0) = 80 \]
Find the time of the peak with the vertex formula
Why: The opposite of sixty-four over twice negative sixteen is two. This answers the when, in seconds.
\[ t = \frac{-64}{2(-16)} = \frac{-64}{-32} = 2 \]
Substitute to get the greatest height
Why: The formula gave a time; the height must be computed. Square the two first, then multiply by negative sixteen.
\[ h(2) = -16(2)^2 + 64(2) + 80 = -64 + 128 + 80 = 144 \]
For the time in the air, set the height to zero
Why: The ball is in the air until it hits the ground, which is a height of zero - not a height of eighty. Dividing through by negative sixteen makes the numbers friendly.
\[ -16t^2 + 64t + 80 = 0 \;\Rightarrow\; t^2 - 4t - 5 = 0 \;\Rightarrow\; (t - 5)(t + 1) = 0 \]
Discard the impossible root
Why: Negative one second is before the throw, which the model does not describe. Only the positive time is meaningful, so the ball lands after five seconds.
\[ t = 5 \quad \text{(reject } t = -1\text{)} \]
Verify the landing time and the symmetry of the flight
Why: The height really is zero at five seconds, and equal times either side of the peak give equal heights. Greatest height one hundred forty-four feet at two seconds; time in the air five seconds.
\[ h(5) = -400 + 320 + 80 = 0, \qquad h(1) = 128 = h(3) \]
Picture it
Animation
Shows: Each line of the worked example "how high, and for how long", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The leading coefficient is negative, so the path peaks and comes back down. The constant term is the height at time zero, which matches the eighty-foot platform.
Ranking
Put in order
These are the steps of Pattern: any maximum-or-minimum word problem, scrambled. Put them back in order before the next slide shows you.
Why: This is the order the recipe itself gives. Recalling the sequence without the slide in front of you is the difference between recognising the method and being able to run it — most of what goes wrong in practice is a step done out of turn.
Pattern
That last step is the fastest insurance in the whole course. If a neighbouring input beats your supposed maximum, something is wrong and you have caught it before the grader did.
Real world
Discussion prompt
Outside this lesson: where does Quadratic Functions, Parabolas, and Optimization actually turn up? Name one concrete situation — a job, a piece of software someone ships, a decision somebody has to make — and say which part of Pattern: any maximum-or-minimum word problem is doing the work in it.
Hint: Vague is the failure mode here. "Engineering" is not a situation; "deciding whether this build is fast enough to ship" is.
Answer:
That deck covers everything a parabola can tell you: vertex form and the sign trap hiding inside it, the vertex formula from standard form, converting between forms by completing the square, intercepts and what the discriminant says about how many there are, domain and range, and applied optimization for maximum area, maximum revenue, and projectile height. It targets the four errors that cost the most points: flipping the sign of the vertex, dropping the negative in the vertex formula, reporting where the maximum happens instead of the maximum value, and assuming that every parabola crosses the axis twice.
Elimination
Eliminate the wrong options
What is the maximum height the ball reaches?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: The vertex time is the opposite of 48 over twice negative 16, which is 48 over 32, or 1.5 seconds. Substituting gives negative 16 times 2.25, plus 48 times 1.5, plus 6, which is negative 36 plus 72 plus 6, or 42 feet.
Check
A ball is thrown from a six-foot ledge. Its height in feet after a number of seconds is given below.
\[ h(t) = -16t^2 + 48t + 6 \]
Check your understanding
What is the maximum height the ball reaches?
Answer: A
Why: The vertex time is the opposite of 48 over twice negative 16, which is 48 over 32, or 1.5 seconds. Substituting gives negative 16 times 2.25, plus 48 times 1.5, plus 6, which is negative 36 plus 72 plus 6, or 42 feet.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — The Shape Itself · Vertex Form: the graph, written down · Standard Form and the Vertex Formula · Completing the Square · Intercepts and the Discriminant · Optimization: the reason any of this matters. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
Almost every parabola question is really the same question: where is the vertex, and which of its two numbers do you need?
| The mistake | The habit that stops it |
|---|---|
| reading the vertex sign straight off vertex form | rewrite a plus inside as minus a negative |
| dropping the minus in the vertex formula | write the minus sign before substituting |
| adding a completing number without paying it back | add and subtract on the same line |
| reporting the input as the maximum value | underline the question word first |
| assuming two crossings | compute the discriminant before solving |
Next up: combining functions, composing them, and running one backwards to build its inverse.
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