Reads a parabola entirely from its formula. Establishes vertex form as the transformation form of the squaring rule, converts to it by completing the square and back by expanding, locates the vertex from standard form directly, and uses the discriminant to predict how many real roots exist. Closes with optimisation, where the vertex answers maximum and minimum questions without calculus.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 3 — Polynomial and Rational Functions
§3.2 Quadratic Functions, pp. 286-310
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §3.2 Quadratic Functions §3.2, pp. 286-310 — the pages these objectives are drawn from
Warm-up
Section 1.5 transformed the squaring rule. This section names what those transformations produce.
Discussion prompt
Without expanding anything, describe the graph of the rule that squares the quantity x minus 3 and then subtracts 5.
Hint: Which transformations from Section 1.5 are these, and what do they do to the lowest point?
Answer:
It is the squaring rule moved right 3 and down 5, so its lowest point has moved from the origin to the point with coordinates 3 and negative 5.
That point is the vertex, and the form the rule is written in shows it directly. Nothing had to be computed, factored or plotted.
So vertex form is not a new idea — it is §1.5's standard transformation form applied to one particular parent. The work of this section is getting a quadratic into that form, because once it is there the graph reads off.
Concept
Written as a constant times a squared bracket plus a constant, a quadratic displays its vertex and its transformations directly. The number subtracted inside is the horizontal shift, and the number added outside is the vertical one.
vertex — The turning point of a parabola: its lowest point when it opens upward and its highest when it opens downward. It lies on the axis of symmetry, and it is the maximum or minimum of the function.
\[ f(x)=a(x-h)^2+k, \qquad \text{vertex } (h,k) \]
The letter a does two jobs at once. Its sign decides which way the parabola opens, and therefore whether the vertex is a maximum or a minimum; its size decides how narrow the parabola is. Both are read off without any computation.
Figure (svg): Standard form and vertex form of a quadratic side by side, each with the information it displays directly highlighted, and the conversion between them named in each direction
OpenStax, Precalculus, §3.2 Quadratic Functions §3.2, pp. 286-291
Section
Section 1
Concept
A parabola is symmetric about a vertical line through its vertex. Every feature — the intercepts, the shape, the maximum or minimum — is positioned relative to that line.
The equidistance in the fourth bullet is worth exploiting. If both x-intercepts are known, the axis is at their average, and that gives the vertex's input without any formula at all — often the fastest route when the quadratic factors easily.
Figure (svg): A parabola with every feature labelled: the vertex, the axis of symmetry as a dashed vertical line, the y-intercept, and the two x-intercepts placed symmetrically about the axis
OpenStax, Precalculus, §3.2 Quadratic Functions §3.2, pp. 286-292
Picture it
One vertical line organises the entire picture.
Figure (svg): A parabola with every feature labelled: the vertex, the axis of symmetry as a dashed vertical line, the y-intercept, and the two x-intercepts placed symmetrically about the axis
The two x-intercepts sit three units either side of the axis. Averaging them recovers the axis, which is the symmetry stated as a computation.
Worked example
No computation required beyond reading the signs.
\[ \text{Describe the graph of } f(x)=-2(x+1)^2+8. \]
Read the vertex
Why: Inside is x minus negative 1, so h is negative 1.
\[ \text{vertex } (-1, 8) \]
Read the opening direction
Why: The leading coefficient is negative.
Read the width
Why: Its size is 2, greater than 1.
Identify the extremum
Why: Opening down makes the vertex the highest point.
\[ \text{maximum value } 8 \]
Figure (svg): The solution to Worked example read the features from vertex form shown as a ladder of expressions, one row per legal move
\[ \text{vertex } (-1,8), \text{ opens down, max } 8, \text{ axis } x=-1 \]
Verify: check the vertex by substitution
Why: At x equal to negative 1 the bracket is zero, so the output is 8 — confirming both coordinates. The sign convention is the §1.5 one: a plus inside means a shift left, so the vertex is at negative 1 rather than 1.
OpenStax, Precalculus, §3.2 Quadratic Functions §3.2, pp. 288-290
Matching
Watch the sign inside the bracket.
Match the pairs
Why: The sign inside reverses to give the input coordinate and the sign outside carries straight through to the output coordinate. The last one also has a leading minus, which flips the opening direction and turns the vertex from a minimum into a maximum without moving it.
Worked example
Symmetry gives the axis for free when the roots are known.
\[ \text{A parabola has } x\text{-intercepts at } -2 \text{ and } 6. \text{ Where is its axis?} \]
Recall the symmetry
Why: The two roots are equidistant from the axis.
Average them
Why: The midpoint is the axis.
\[ \frac{-2 + 6}{2} \]
Compute
Why: Four over two.
\[ x = 2 \]
Note what follows
Why: The vertex sits on that line.
\[ \text{vertex input is } 2 \]
Figure (svg): The solution to Worked example find the vertex from the intercepts shown as a ladder of expressions, one row per legal move
\[ x = \frac{-2+6}{2} = 2 \]
Verify: check the distances
Why: From 2 to negative 2 is 4, and from 2 to 6 is also 4. The two roots really are equidistant, as symmetry requires. Finding the output still needs the actual function, but the input is settled by the roots alone — which is often the quicker half.
OpenStax, Precalculus, §3.2 Quadratic Functions §3.2, pp. 291-292
Trap
\[ f(x)=(x+4)^2-1 \;\Longrightarrow\; \text{vertex } (4,-1) \]
Read the two constants from the formula
Why: The 4 appears inside and the negative 1 outside, so they are taken as the coordinates.
The vertex is reported as the point with input 4.
The form subtracts h, so a plus 4 inside means h is negative 4. The vertex is at the point with input negative 4.
This is the §1.5 sign reversal, unchanged: inside changes are horizontal and reversed.
Check by asking which input makes the bracket zero. Here x plus 4 vanishes at negative 4, and that is where the vertex is — a one-line check that never depends on remembering the convention.
Prediction
A parabola opens upward and its vertex sits above the horizontal axis.
Predict first
How many x-intercepts does it have?
Correct: None, since the whole curve is above the axis.
Why: An upward parabola has its vertex as its lowest point, so if that lowest point is above the axis the curve never reaches it. One intercept would require the vertex to sit exactly on the axis, and two would require it below. The vertex's position relative to the axis decides the count entirely.
Faded example
A parabola crosses the horizontal axis at negative 5 and at 1.
Fill in the blanks
x = \frac-23 = ___, \qquad \text___ = ___
Why: The average of the two roots is negative 2, which is the axis. Each root is 3 units away, confirming the symmetry. This route to the vertex's input needs no formula and no completing the square, and it is the fastest when the roots are already known.
Explain it to yourself
A parabola is symmetric about its axis, and the roots reflect that.
Discussion prompt
Explain why the two x-intercepts of a parabola must be equidistant from its vertex.
Hint: What does the reflection do to a point on the curve at height zero?
Answer:
Reflecting the graph in its axis leaves it unchanged, so every point on the curve has a mirror partner also on the curve, at the same height and the same distance the other side of the axis.
A point at height zero is an x-intercept, so its mirror partner is also at height zero and is also an x-intercept. The two therefore sit at equal distances from the axis, one each side.
The one exception is when the point lies on the axis, which is its own mirror image. That is the repeated-root case, where the two intercepts have merged into one at the vertex — which is exactly the discriminant-equal-to-zero case later in this lesson.
Section
Section 2
Concept
To convert to vertex form, add the number that turns the quadratic and linear terms into a perfect square, and subtract the same number to leave the value unchanged.
The fourth and fifth bullets are where most errors live. Factoring out a leading coefficient of 3 means the number added inside the bracket is worth three times as much outside it, and forgetting that scaling gives a vertex at the wrong height.
Figure (svg): The algebraic steps of completing the square shown as an area diagram: a square of side x, two rectangles of width half the linear coefficient, and the small square that must be added to complete the figure
OpenStax, Precalculus, §3.2 Quadratic Functions §3.2, pp. 292-298
Picture it
The square and the two rectangles leave a corner missing, and its area is what must be added.
Figure (svg): The algebraic steps of completing the square shown as an area diagram: a square of side x, two rectangles of width half the linear coefficient, and the small square that must be added to complete the figure
Half of 6 is 3, and the missing corner is 3 by 3. That is the whole rule, and the picture explains why it is half and why it is squared.
Worked example
The simplest case, where nothing has to be factored out.
\[ \text{Write } f(x)=x^2-8x+11 \text{ in vertex form.} \]
Halve the linear coefficient and square
Why: Half of negative 8 is negative 4; its square is 16.
\[ \text{the number is } 16 \]
Add and subtract it
Why: Keeping the value unchanged.
\[ x ^{2} - 8 x + 16 - 16 + 11 \]
Group the perfect square
Why: The first three terms factor.
\[ (x - 4) ^{2} - 16 + 11 \]
Collect the constants
Why: Negative 16 plus 11.
\[ (x - 4) ^{2} - 5 \]
Figure (svg): The algebraic steps of completing the square shown as an area diagram: a square of side x, two rectangles of width half the linear coefficient, and the small square that must be added to complete the figure
\[ f(x)=(x-4)^2-5, \quad \text{vertex } (4,-5) \]
Verify: expand back and compare
Why: Expanding gives x squared minus 8x plus 16, minus 5, which is x squared minus 8x plus 11 — the original. Expanding back is the reliable check on any completed square, and it catches both arithmetic slips and sign errors in one step.
OpenStax, Precalculus, §3.2 Quadratic Functions §3.2, pp. 293-295
Faded example
Complete the square on the expression x squared minus 10x.
Fill in the blanks
\text-5 -10 \text25 ___; \text___ ___ \;\Longrightarrow\; (x-5)^2 - 25
Why: Half of negative 10 is negative 5, and squaring gives positive 25. The bracket carries the negative 5, and the 25 is subtracted outside to compensate for having been added. Note that the number added is always positive, since it is a square, however the linear coefficient was signed.
Worked example
Factor it out of the x terms first, and watch what happens on the way back.
\[ \text{Write } f(x)=3x^2+12x+7 \text{ in vertex form.} \]
Factor 3 from the x terms only
Why: The constant stays outside.
\[ 3(x ^{2} + 4 x) + 7 \]
Complete the square inside
Why: Half of 4 is 2; its square is 4.
\[ 3(x ^{2} + 4 x + 4 - 4) + 7 \]
Take the subtracted 4 outside, multiplied by 3
Why: This is the step that is skipped.
\[ 3(x + 2) ^{2} - 12 + 7 \]
Collect
Why: Negative 12 plus 7.
\[ 3(x + 2) ^{2} - 5 \]
Figure (svg): The solution to Worked example a leading coefficient other than 1 shown as a ladder of expressions, one row per legal move
\[ f(x)=3(x+2)^2-5, \quad \text{vertex } (-2,-5) \]
Verify: check the vertex numerically
Why: At x equal to negative 2 the original gives 12 minus 24 plus 7, which is negative 5 — matching. The 4 that was subtracted inside the bracket came out as 12 because it was inside a factor of 3, and forgetting that multiplication would have put the vertex at height 3 instead of negative 5.
OpenStax, Precalculus, §3.2 Quadratic Functions §3.2, pp. 296-298
Error analysis
A student completes the square with a leading coefficient of 2.
Annotate
On: \( 2x^2+8x+3 = 2(x^2+4x+4) - 4 + 3 = 2(x+2)^2 - 1 \)
Whatever is added inside a bracket is worth the leading coefficient times as much outside it. Expanding the finished answer back is the check that catches this every time.
Prediction
A quadratic has leading coefficient 5, and completing the square inside the bracket requires adding 9.
Predict first
What must be subtracted outside?
Correct: 45, because the 9 was inside a factor of 5.
Why: The bracket is multiplied by 5, so adding 9 inside adds 45 to the expression's value. Subtracting only 9 would leave the expression 36 larger than it started, putting the vertex at the wrong height. This scaling is the single most common error in completing the square.
Sorting
The constant must be the square of half the linear coefficient.
Sort into buckets
Sort each expression.
Socratic
The rule is stated as half the coefficient, squared.
Discussion prompt
Explain where both operations come from, using the expansion of a squared bracket.
Hint: What is the middle term when you expand x plus p, all squared?
Answer:
Expanding x plus p, all squared, gives x squared plus 2px plus p squared. So the linear coefficient of a perfect square is always twice the number in the bracket.
Running that backwards: given a linear coefficient, the number in the bracket is half of it. That is where the halving comes from — it undoes the doubling in the expansion.
And the constant term of the expansion is that number squared, which is what must be present for the trinomial to be a perfect square. So both operations are simply the expansion read in reverse, and the area picture shows the same thing geometrically.
Section
Section 3
Concept
Completing the square on the general quadratic gives the vertex's input directly. It is negative b over 2a, and the output is found by substituting it back.
\[ h = -\frac{b}{2a}, \qquad k = f(h) \]
The connection with the quadratic formula is worth seeing. That formula gives the roots as negative b over 2a, plus or minus something, so the two roots are symmetric about negative b over 2a — which is exactly the statement that the axis passes through their midpoint.
Figure (svg): A parabola with every feature labelled: the vertex, the axis of symmetry as a dashed vertical line, the y-intercept, and the two x-intercepts placed symmetrically about the axis
OpenStax, Precalculus, §3.2 Quadratic Functions §3.2, pp. 298-302
Picture it
The formula locates the dashed line, and the roots sit symmetrically about it.
Figure (svg): A parabola with every feature labelled: the vertex, the axis of symmetry as a dashed vertical line, the y-intercept, and the two x-intercepts placed symmetrically about the axis
The quadratic formula's structure says the same thing: a centre at negative b over 2a, with the radical measuring how far each root sits from it.
Worked example
One formula, then one substitution.
\[ \text{Find the vertex of } f(x)=2x^2-12x+7. \]
Identify a and b
Why: The coefficients of the squared and linear terms.
\[ a = 2, b = -12 \]
Apply the formula
Why: Negative b over 2a.
\[ \frac{12}{4} = 3 \]
Substitute back for the output
Why: Into the original rule.
\[ 2(9) - 36 + 7 \]
Compute
Why: Eighteen minus thirty six plus seven.
\[ -11 \]
Figure (svg): The solution to Worked example find the vertex without completing the square shown as a ladder of expressions, one row per legal move
\[ \text{vertex } (3,-11) \]
Verify: check against completing the square
Why: Factoring 2 out gives 2 times the quantity x squared minus 6x, plus 7; completing gives 2 times the square of x minus 3, minus 18, plus 7, which is minus 11. Same vertex, more steps — which is exactly why the formula is worth having when only the vertex is needed.
OpenStax, Precalculus, §3.2 Quadratic Functions §3.2, pp. 299-300
Faded example
Find the axis of symmetry of the rule 3x squared plus 18x minus 2.
Fill in the blanks
h = -\frac6-3 = -\frac______} = ___
Why: The denominator is twice the leading coefficient, which is 6, and the minus sign in the formula makes the answer negative 3. Substituting back into the original gives the output, completing the vertex. The sign is the part worth double-checking, since the formula's minus is easy to drop.
Worked example
A second route when the quadratic factors easily.
\[ \text{Find the vertex of } f(x)=x^2-2x-15. \]
Factor
Why: Two numbers multiplying to negative 15, adding to negative 2.
\[ (x - 5) (x + 3) \]
Read the roots
Why: Where each factor vanishes.
\[ x = 5\text{ and } x = -3 \]
Average them
Why: The axis sits at their midpoint.
\[ \frac{5 + (-3)}{2} = 1 \]
Substitute for the output
Why: Into the original rule.
\[ 1 - 2 - 15 = -16 \]
Figure (svg): The solution to Worked example the vertex is the average of the roots shown as a ladder of expressions, one row per legal move
\[ \text{vertex } (1,-16) \]
Verify: confirm with the formula
Why: Negative b over 2a gives 2 over 2, which is 1 — the same axis. Both routes agree, and the factoring route also delivered the two intercepts for free, which the vertex formula does not. Which route is faster depends on whether the quadratic factors nicely.
OpenStax, Precalculus, §3.2 Quadratic Functions §3.2, pp. 300-302
Trap
\[ f(x)=x^2+6x+5: \quad h = \frac{b}{2a} = \frac{6}{2} = 3 \]
Apply the formula as recalled
Why: The coefficients are identified correctly and the division carried out.
The axis is reported as x equal to 3.
The formula has a minus sign: it is negative b over 2a, giving negative 3.
Check it by factoring: the rule factors as x plus 5 times x plus 1, with roots at negative 5 and negative 1, whose average is negative 3. The sign matters.
A quick sanity check: an upward parabola with a positive b opens to the right of its vertex, so the vertex sits at a negative input. Comparing the answer's sign against the sign of b catches this immediately.
Prediction
A quadratic has a positive leading coefficient and a positive linear coefficient.
Predict first
Is its vertex to the left or the right of the vertical axis?
Correct: To the left, since -b/(2a) is negative.
Why: With both a and b positive, negative b over 2a is a negative number, so the vertex sits at a negative input. The constant term shifts the parabola vertically and has no effect at all on where the axis is, which is a useful thing to notice: the axis depends only on a and b.
Matching
Three ways to find a vertex, with different by-products.
Match the pairs
Why: All three give the same vertex. Completing the square additionally produces vertex form, which shows every transformation. Averaging the roots requires factoring first, which hands you the intercepts. The formula is the fastest and the least informative, which makes it the right choice when the vertex is all that is wanted.
Explain it to yourself
Negative b over 2a is not an arbitrary expression.
Discussion prompt
Explain its connection to the quadratic formula.
Hint: Write the quadratic formula and look at the part outside the radical.
Answer:
The quadratic formula gives the roots as negative b over 2a, plus or minus the radical over 2a. The first part is the same expression as the vertex formula.
So the two roots sit symmetrically either side of negative b over 2a, at equal distances given by the radical term. Their average is therefore exactly negative b over 2a.
That is the axis of symmetry, arrived at from the roots rather than from completing the square. The two derivations agree because the quadratic formula is itself obtained by completing the square in general — so the vertex formula is a fragment of that same computation.
Section
Section 4
Concept
The quantity under the radical in the quadratic formula decides how many real roots exist, before any solving is attempted.
\[ \Delta = b^2-4ac \]
The last bullet is a practical time-saver. Computing the discriminant before choosing a method tells you whether factoring is likely to work: a perfect square means it will, and anything else means the quadratic formula or completing the square is the sensible route.
Figure (svg): Three parabolas showing the three cases of the discriminant: two real roots crossing the axis twice, one repeated root touching it, and no real roots clearing it entirely
OpenStax, Precalculus, §3.2 Quadratic Functions §3.2, pp. 302-306
Picture it
The sign of one number decides which of these pictures you have.
Figure (svg): Three parabolas showing the three cases of the discriminant: two real roots crossing the axis twice, one repeated root touching it, and no real roots clearing it entirely
With complex numbers from §3.1 available, all three cases have two roots. The discriminant now reports what kind rather than how many.
Worked example
One computation saves choosing the wrong method.
\[ \text{Describe the roots of } 2x^2-4x+5=0 \text{ without solving.} \]
Identify the coefficients
Why: In order.
\[ a = 2, b = -4, c = 5 \]
Compute the discriminant
Why: b squared minus 4ac.
\[ 16 - 40 \]
Read the sign
Why: It is negative.
\[ -24 \]
State the conclusion
Why: No real roots; two complex conjugates.
Figure (svg): Three parabolas showing the three cases of the discriminant: two real roots crossing the axis twice, one repeated root touching it, and no real roots clearing it entirely
\[ \Delta = -24 < 0: \text{ two complex conjugate roots} \]
Verify: confirm from the graph
Why: The leading coefficient is positive so the parabola opens up, and its vertex is at input 1 with output 3 — above the axis. An upward parabola whose lowest point is above the axis never meets it, so there are indeed no real roots, which agrees with the discriminant.
OpenStax, Precalculus, §3.2 Quadratic Functions §3.2, pp. 303-305
Sorting
Compute the discriminant and read its sign.
Sort into buckets
Sort each quadratic.
Worked example
The boundary case, where the two roots have merged.
\[ \text{Solve } x^2-6x+9=0. \]
Compute the discriminant
Why: Thirty six minus thirty six.
\[ 0 \]
Predict the outcome
Why: A zero discriminant means one repeated root.
Factor
Why: It is a perfect square trinomial.
\[ (x - 3) ^{2} = 0 \]
Solve
Why: The only root.
\[ x = 3,\text{ twice} \]
Figure (svg): The solution to Worked example a repeated root shown as a ladder of expressions, one row per legal move
\[ x = 3 \text{ (repeated)} \]
Verify: check the graph's behaviour there
Why: The vertex is at input 3 with output zero, so the parabola touches the axis without crossing. That is what a repeated root looks like: the curve comes down to the axis and turns back rather than passing through, which §3.4 will connect to even multiplicity in general.
OpenStax, Precalculus, §3.2 Quadratic Functions §3.2, pp. 305-306
Trap
\[ x^2-5x+6: \quad \Delta = (-5)^2-4(1)(6) \overset{?}{=} -25-24 = -49 \]
Substitute into the discriminant formula
Why: The three coefficients are identified correctly.
The square of negative 5 is written as negative 25, giving a negative discriminant and no real roots.
A negative squared is positive. Negative 5 squared is 25, so the discriminant is 25 minus 24, which is 1.
A positive discriminant means two real roots, and indeed the quadratic factors as x minus 2 times x minus 3.
Bracket the coefficient before squaring, as in §1.1's substitution rule. The b term is squared, so its sign never survives, and a negative discriminant arising from a squared term is always an error.
Prediction
A quadratic with two real roots has its constant term increased steadily.
Predict first
What happens to the discriminant?
Correct: It decreases, eventually becoming negative.
Why: The discriminant subtracts 4ac, so increasing c with a positive reduces it. Geometrically, raising the constant lifts the whole parabola: the two roots move together, merge when the discriminant hits zero and the vertex touches the axis, then vanish as it lifts clear. The three cases are one continuous process.
Faded example
Find the discriminant of 3x squared minus 2x plus 5.
Fill in the blanks
\Delta = (-2)^2 - 4(3)(5) = 4 - 60 = -56
Why: Squaring negative 2 gives positive 4, and 4 times 3 times 5 is 60. The result is negative 56, so there are no real roots and the parabola clears the axis. The sign of b never survives the squaring, which is the check that catches the standard error.
Edge cases
The discriminant sits exactly at zero for one particular value of the constant term.
Discussion prompt
What is happening to the graph at that instant, and what happens either side of it?
Hint: Where are the two roots as the discriminant approaches zero?
Answer:
As the discriminant shrinks towards zero the two roots move towards each other, and at zero they merge at a single point — the vertex, which is sitting exactly on the axis.
Just before that, there are two distinct crossings very close together. Just after, the vertex lifts clear and there are no real roots at all, though the two complex roots are there, having moved off the real axis as a conjugate pair.
So the three cases are not three separate situations but one continuous process viewed at three moments. That is worth knowing because it explains why the boundary case is so often the interesting one in applications: it is the threshold where a solution appears or disappears.
Section
Section 5
Concept
When a quantity to be maximised or minimised can be written as a quadratic, the vertex gives the answer directly — both where the extremum occurs and what its value is.
The first step is the one that takes thought. A problem about a rectangle gives two dimensions and one constraint, and the constraint is used to express the second dimension in terms of the first, so that the area becomes a function of one variable. Everything after that is this lesson.
Figure (svg): A rectangle with a fixed perimeter, its area plotted against one side length as a downward parabola, with the maximum marked at the vertex
OpenStax, Precalculus, §3.2 Quadratic Functions §3.2, pp. 306-310
Picture it
The constraint turned a two-variable problem into a one-variable quadratic.
Figure (svg): A rectangle with a fixed perimeter, its area plotted against one side length as a downward parabola, with the maximum marked at the vertex
The domain runs from 0 to 20, since a width outside that gives a negative dimension. The maximum sits comfortably inside it, at the vertex.
Worked example
Use the constraint to eliminate one variable, then find the vertex.
\[ \text{A rectangle has perimeter } 40. \text{ Maximise its area.} \]
Write the constraint
Why: Twice the sum of the sides is 40.
\[ w + l = 20 \]
Express one side via the other
Why: Solve the constraint.
\[ l = 20 - w \]
Write the area as one variable
Why: Width times length.
\[ A(w) = w(20 - w) \]
Find the vertex
Why: Expand and apply the formula.
\[ w = 10, A = 100 \]
Figure (svg): A rectangle with a fixed perimeter, its area plotted against one side length as a downward parabola, with the maximum marked at the vertex
\[ w = 10, \; l = 10, \; A_{\max} = 100 \]
Verify: check the shape and the domain
Why: The leading coefficient of the expanded area is negative 1, so the parabola opens down and the vertex is a maximum, as required. The width 10 lies inside the allowed domain from 0 to 20. Note the answer is a square, which is the general result for a fixed perimeter.
OpenStax, Precalculus, §3.2 Quadratic Functions §3.2, pp. 307-309
Sorting
The vertex has two coordinates, and questions ask for one or the other.
Sort into buckets
Sort each question.
Worked example
Price and quantity are linked, and revenue is their product.
\[ \text{At } 30 \text{ per ticket } 500 \text{ sell; each } 1 \text{ rise loses } 10 \text{ sales. Maximise revenue.} \]
Let x be the number of one-dollar rises
Why: This is the variable to optimise over.
\[ x =\text{ number of rises} \]
Write the price and the quantity
Why: Each depends on x linearly.
\[ \text{price } 30 + x,\text{ sales } 500 - 10 x \]
Multiply for revenue
Why: Revenue is price times quantity.
\[ R(x) = (30 + x) (500 - 10 x) \]
Find the vertex
Why: Expand and apply the formula.
\[ x = 10 \]
Figure (svg): The solution to Worked example maximise revenue shown as a ladder of expressions, one row per legal move
\[ x = 10: \text{ price } 40, \text{ sales } 400, \text{ revenue } 16\,000 \]
Verify: check against the starting revenue
Why: At the original price, revenue was 30 times 500, which is 15000. At the optimum it is 40 times 400, which is 16000 — genuinely higher, so the answer improves on the starting point as it should. Note also that raising the price further to 50 gives 50 times 300, which is 15000 again: the vertex really is the peak.
OpenStax, Precalculus, §3.2 Quadratic Functions §3.2, pp. 309-310
Error analysis
A student maximises an area and reports the answer.
Annotate
On: \( A(w)=w(20-w), \; \text{vertex at } w=10 \;\Longrightarrow\; \text{maximum area is } 10 \)
This is §1.3's location-versus-value distinction, in an applied setting. Optimisation questions usually want both numbers, and reading which is asked for is worth doing carefully.
Prediction
A quantity is modelled by a quadratic with a positive leading coefficient.
Predict first
Does its vertex give a maximum or a minimum?
Correct: A minimum, since it opens upward.
Why: A positive leading coefficient means the parabola opens upward, so the vertex is its lowest point. A maximum requires a negative leading coefficient. Checking the sign of a before reporting is worth doing, because an optimisation question asking for a maximum and a model that opens upward is a sign the model was set up wrongly.
Faded example
A rectangular pen uses 60 metres of fence with one side against a wall, so only three sides are fenced.
Fill in the blanks
2w + l = 60 \;\Longrightarrow\; l = 60 - 2w, \qquad A(w) = w(60 - ___w)
Why: Two widths and one length are fenced, so the constraint is 2w plus l equal to 60, giving l as 60 minus 2w. The area is then w times that. The wall changes the constraint and therefore the answer: the optimum here is a width of 15 and a length of 30, which is not a square.
Real world
Optimisation is one of the most-used ideas in the whole course.
Discussion prompt
Why is being able to optimise a quadratic without calculus worth having, given that Chapter 12 will do it more generally?
Hint: How many real optimisation problems are quadratic?
Answer:
A great many practical optimisation problems are quadratic: area under a constraint, revenue when demand falls linearly with price, projectile height. For all of these the vertex is the complete answer and no calculus is required.
Calculus generalises the method to any differentiable function, which is a genuine advance. But it arrives at the same answer for a quadratic by a longer route — setting the derivative to zero gives 2ax plus b equal to zero, which solves to negative b over 2a, exactly this section's formula.
So the vertex formula is not a stopgap superseded later; it is the quadratic case of the general result, and it remains the fastest route whenever the model happens to be quadratic — which it often is.
Comparison
Fill the blanks from memory. Choosing the right form is most of the efficiency in this section.
Comparison matrix
| standard form | vertex form | |
|---|---|---|
| written as | ax^2 + bx + c | a(x - h)^2 + k |
| shows for free | the y-intercept, c | the vertex, (h, k) |
| vertex found by | the formula -b/(2a) | reading it off |
| best for | factoring and the quadratic formula | graphing and optimisation |
| converted to the other by | completing the square | expanding |
The conversion is one-way easy: expanding is mechanical, completing the square takes care. That asymmetry is why the vertex formula exists — it extracts the one fact you usually want without doing the harder conversion.
Pattern
Six steps, and after them nothing about the parabola is unknown.
Step 5 before step 6 is deliberate: knowing the number of real roots tells you whether to attempt factoring, use the quadratic formula, or stop.
OpenStax Algebra and Trigonometry 2e, §5.1 Quadratic Functions §5.1
Check
Watch the sign inside.
Check your understanding
What is the vertex of the parabola given by 2(x + 5)^2 - 3?
Answer: A
Why: The bracket vanishes at x equal to negative 5, so that is the vertex's input, and the constant outside gives the output as negative 3. The plus inside reverses to a negative coordinate, as inside changes always do.
Check
Half, then square.
Check your understanding
What number completes the square on x squared plus 14x?
Answer: A
Why: Half of 14 is 7, and 7 squared is 49. The expression becomes the square of x plus 7, minus 49 if the value is to be preserved.
Check
Square the b, sign and all.
Check your understanding
How many real roots does 4x squared minus 12x plus 9 have?
Answer: A
Why: The discriminant is 144 minus 144, which is zero, so there is one repeated root. The quadratic is a perfect square, factoring as the square of 2x minus 3, and the parabola touches the axis at its vertex without crossing.
Real world
Anything thrown, dropped or launched follows a parabola, and the vertex is the part everyone wants.
Discussion prompt
A ball's height is modelled by a quadratic in time. What do the vertex, the roots and the y-intercept each mean?
Hint: What is happening at the highest point, at ground level, and at the start?
Answer:
The vertex is the highest point of the flight: its input is the time at which the peak is reached and its output is the maximum height. Both are usually what a question wants.
The roots are the times at which the height is zero, which is ground level. The positive one is when it lands; the negative one is usually outside the model's domain and has no physical meaning.
The y-intercept is the height at time zero, which is the height it was launched from. Note how §2.3's domain discussion applies here unchanged: the model is meaningful from the launch until the landing, and the parabola continues on paper well past both.
Commit first
State your confidence along with your answer.
Predict first
A parabola opens downward and its vertex lies below the horizontal axis. How many x-intercepts does it have?
Correct: None, since its highest point is below the axis.
Why: A downward parabola has its vertex as its maximum, so if even that is below the axis then nothing on the curve reaches it. The discriminant would be negative. The size of the leading coefficient changes the width but cannot lift any point above the vertex, so it has no bearing on the count.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
Explain to a classmate why completing the square works, using the area picture rather than the rule.
Hint: Draw a square of side x with two rectangles attached.
Answer:
Draw a square of side x, then attach two rectangles along two of its sides, each of width half the linear coefficient. Together those rectangles account for the whole linear term, split evenly.
The figure is now a large square with one corner missing, and that corner is a small square whose side is the same half-coefficient. Its area is therefore that number squared — which is exactly what the rule says to add.
The subtraction afterwards is bookkeeping: you added area that was not there, so you take the same amount away to keep the expression's value unchanged. The picture explains both the halving and the squaring, which a memorised rule does not.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The second is the most error-prone piece of algebra in the section, and the scaling step with a leading coefficient is where nearly all of those errors live. The fourth is the least mechanical: the algebra is easy once the set-up is right, and the set-up is the part that takes thought.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Draw one parabola and label its vertex, axis of symmetry, y-intercept and any x-intercepts. Beside it, write the same function in both standard and vertex form, and show the completing-the-square steps that connect them. Underneath, write the discriminant of your function and say which of the three cases it falls into.
If your two forms expand into each other and your discriminant's sign matches the number of x-intercepts you drew, every part of the section is consistent on your page.
Recap
Five things, and the first one makes §1.5 pay for itself.
| if you remember one thing | it should be this |
|---|---|
| about vertex form | it is Section 1.5's transformation form for the squaring rule |
| about completing the square | half the coefficient, squared - and scale the compensation |
| about the discriminant | its sign counts the real roots before you solve |
| about optimisation | the vertex's input is where, and its output is what |
Section 3.3 leaves degree two behind and asks what happens as the degree grows, where the end behaviour and the number of turning points both become questions about the leading term.
OpenStax, Precalculus, §3.2 Quadratic Functions §3.2, pp. 286-310 — everything on these slides traces back here
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