The quadratic parent function and its parabola, vertex and axis of symmetry, how a controls width and direction and c controls height, the vertex formula for the general standard form, and finding a minimum or maximum value in a revenue model.
Subject: Algebra 2 · 65 slides · symbolic lesson
Open the interactive version of this deck
Title
Algebra 2 · Chapter 4 — Quadratic Functions and Factoring
Graph Quadratic Functions in Standard Form
Objectives
Five outcomes. The fourth is the formula the rest of the chapter leans on.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.1 Graph Quadratic Functions in Standard Form §4.1, pp. 236-241 — the lesson these objectives are drawn from
Warm-up
Lesson 2.7 took the parent V and moved it around with a, h and k. This chapter does the same with a new parent.
Discussion prompt
Evaluate x squared at x equal to -3, -1, 0, 1 and 3. Plot those five points in your head. How does the shape differ from the V of the absolute value function?
Hint: Compare the gap between consecutive heights.
Answer:
\[ 9, \; 1, \; 0, \; 1, \; 9 \]
Still symmetric about the vertical axis, but the heights grow much faster further out — 9 rather than 3 at x equal to 3. That accelerating growth is what curves the graph, so the parabola has no straight branches and no sharp corner.
Concept
A quadratic function is one that can be written as y equals a x squared plus b x plus c with a not zero, and its graph is a parabola. As with every family in this book, the questions are the same: where is it, which way does it open, and how wide is it?
parabola — The U-shaped curve that is the graph of every quadratic function. Its highest or lowest point is called the vertex.
\[ y = ax^2 + bx + c, \quad a \neq 0 \]
The parent is f of x equals x squared, with vertex at the origin and axis of symmetry the vertical axis.
Figure (svg): The parabola y equals x squared with its vertex at the origin and its axis of symmetry marked
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.1 Graph Quadratic Functions in Standard Form §4.1, pp. 236-236 — Parent Function for Quadratic Functions
Section
Section 1
Concept
The graph of x squared is a parabola whose lowest point is the origin. A vertical line through the vertex divides it into mirror images, so every point on one side has a partner at the same height on the other.
axis of symmetry — The vertical line through the vertex that divides a parabola into two mirror-image halves.
\[ f(x) = x^2 \]
The symmetry is worth exploiting: every point you plot gives you a second point free, reflected across the axis.
Figure (svg): The parabola y equals x squared with its vertex at the origin and its axis of symmetry marked
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.1 Graph Quadratic Functions in Standard Form §4.1, pp. 236-236
Picture it
The parent function and its symmetry.
Figure (svg): The parabola y equals x squared with its vertex at the origin and its axis of symmetry marked
The points (-2, 4) and (2, 4) are reflections of each other. Because squaring a number and squaring its opposite give the same result, that pairing holds at every height.
Worked example
Five values, and the shape follows.
\[ \text{Graph } f(x) = x^2 \text{ from a table of five values.} \]
Evaluate at two negative inputs
Why: Squaring removes the sign, so negative three gives nine and negative one gives one.
\[ (-3, 9), (-1, 1) \]
Evaluate at zero and two positive inputs
Why: Zero squared is zero; one and three give one and nine.
\[ (0, 0), (1, 1), (3, 9) \]
Plot and join with a SMOOTH curve
Why: The heights grow faster further out, so the graph bends rather than running straight.
Mark the vertex and the axis
Why: The lowest point is the origin, and the vertical axis mirrors the two halves.
\[ \text{vertex } (0, 0),\text{ axis } x = 0 \]
Figure (svg): The solution to Worked example build the parent from a table shown as a ladder of expressions, one row per algebraic move
\[ f(x) = x^2 \]
Verify: check the symmetry on two pairs
Why: The points (1,1) and (-1,1) have equal heights, and so do (3,9) and (-3,9). Squaring a number and its opposite gives the same value, so the mirror property holds at every x — which is why the axis of symmetry is exactly the vertical axis here.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.1 Graph Quadratic Functions in Standard Form §4.1, pp. 236-236
Sorting
Test each point against y equals x squared.
Sort into buckets
Sort each point.
The parent parabola never dips below the axis, for the same reason the parent V never did: squaring, like taking an absolute value, cannot produce a negative.
Worked example
Comparing the growth of x squared with the growth of the absolute value of x.
\[ \text{Compare } x^2 \text{ and } \lvert x \rvert \text{ at } x = 1, 2, 3, 4. \]
Tabulate the absolute value
Why: One, two, three, four — rising by exactly one each time.
\[ 1, 2, 3, 4 \]
Tabulate the square
Why: One, four, nine, sixteen — rising by three, then five, then seven.
\[ 1, 4, 9, 16 \]
Compare the differences
Why: Constant for the V, increasing for the parabola.
Draw the conclusion
Why: A constant rate of change gives a straight branch; a growing one gives a branch that bends upward.
Figure (svg): The solution to Worked example why the branches curve shown as a ladder of expressions, one row per algebraic move
\[ \Delta(x^2) = 3, 5, 7 \qquad \Delta\lvert x \rvert = 1, 1, 1 \]
Verify: look at the differences of the differences
Why: The gaps between the parabola's differences are 5 minus 3 and 7 minus 5, both equal to 2. A constant SECOND difference is the signature of a quadratic, exactly as a constant first difference was the signature of a line in Lesson 2.1.
Trap
\[ f(x) = x^2 \text{ at } x = -2, -1, 0, 1, 2 \]
Plot the five points and join them with straight lines
Why: The plotting technique from Lesson 2.1 is applied without asking whether the graph is straight.
The resulting shape has corners at every plotted point, and it dips below the true curve between them.
\[ f(x) = x^2 \]
Join the points with a SMOOTH curve
Why: The differences between consecutive outputs grow, so the graph bends continuously rather than changing direction at points.
\[ \text{at } x = 1.5: \; x^2 = 2.25, \text{ not } 2.5 \]
Checking a half-way value settles it: the straight segment from (1,1) to (2,4) would pass through 2.5 at x equal to 1.5, but the true value is 2.25 — the curve sags below the chord.
Prediction
Commit before reasoning.
Predict first
What is the range of the parent function f of x equals x squared?
Correct: All non-negative real numbers — zero and everything above it.
\[ \text{domain: all reals} \qquad \text{range: } y \geq 0 \]
Why: A square is never negative, so nothing below zero is reachable, but zero itself is attained at x equal to 0. The range is therefore y at least zero, exactly as it was for the absolute value function in Lesson 2.7. Perfect squares are only the outputs of INTEGER inputs; the square of 1.5 is 2.25, which is in the range and is not a perfect square.
Matching
Four pieces of vocabulary for one curve.
Match the pairs
Why: The condition that a is not zero matters: if a were zero the x squared term would vanish and the function would be linear, with a straight graph and no vertex at all. That is the same exclusion that kept the constant of variation nonzero in Lesson 2.5.
Explain it to yourself
The parent V had a sharp corner; the parent parabola does not.
\[ f(x) = x^2 \quad \text{versus} \quad f(x) = \lvert x \rvert \]
Discussion prompt
Explain why the parabola turns smoothly at its vertex while the V turns sharply. What is different about how the two functions' rates of change behave near zero?
Hint: Compute each function's change over the interval from -0.1 to 0 and from 0 to 0.1.
Answer:
The V's slope jumps straight from negative one to positive one as you cross zero, with nothing in between — that jump is the corner.
The parabola's slope passes through every value in between: it is close to negative 0.2 just left of zero, exactly 0 at the vertex, and close to 0.2 just right. A slope that changes continuously produces a smooth turn.
Section
Section 2
Concept
In y equals a x squared, the size of a decides how narrow the parabola is and the sign of a decides which way it opens. The vertex stays at the origin throughout.
\[ y = ax^2 \]
Narrower when the size of a exceeds one, wider when it is less than one, and opening downward when a is negative — the same three rules as for the V in Lesson 2.7.
Figure (svg): Three parabolas with different values of a: narrower, the parent, and wider and opening downward
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.1 Graph Quadratic Functions in Standard Form §4.1, pp. 236-237 — Graph a function of the form y = ax^2
Picture it
Example 1 and Example 2's coefficients, on one set of axes.
Figure (svg): Three parabolas with different values of a: narrower, the parent, and wider and opening downward
Doubling the coefficient doubles every height, which makes the curve rise faster and therefore look thinner. Narrow and stretched are the same thing seen two ways.
Worked example
Example 1. Compare with the parent at every step.
\[ \text{Graph } y = 2x^2 \text{ and compare it with } y = x^2. \]
Tabulate five values
Why: At negative two the square is four and doubling gives eight; at negative one, two; at zero, zero.
\[ 8, 2, 0, 2, 8 \]
Plot and join smoothly
Why: The symmetry means the right half mirrors the left.
Compare the vertex and axis with the parent
Why: Both are the origin and the vertical axis, unchanged.
Compare the width
Why: Every height is twice the parent's, so the curve rises faster and looks narrower.
Figure (svg): The solution to Worked example a narrower parabola shown as a ladder of expressions, one row per algebraic move
\[ y = 2x^2 \quad \text{is narrower than } y = x^2 \]
Verify: compare the two at one input
Why: At x equal to 2 the parent gives 4 and this gives 8. Being twice as high at the same horizontal distance is exactly what narrower means — the curve reaches any given height sooner.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.1 Graph Quadratic Functions in Standard Form §4.1, pp. 236-236
Sorting
Compare each with the parent parabola.
Sort into buckets
Sort each function by its width.
Two separate questions again: the size gives the width, the sign gives the direction. Answering them one at a time makes both easy.
Worked example
Guided Practice 1. Both effects of a at once.
\[ \text{Graph } y = -4x^2 \text{ and compare it with } y = x^2. \]
Read the size of a
Why: Four, which exceeds one, so the parabola is narrower than the parent.
Read the sign of a
Why: Negative, so it opens downward and the vertex is a maximum.
Tabulate a few values
Why: At plus or minus one the value is negative four; at plus or minus two, negative sixteen.
\[ (1, -4), (2, -16) \]
Plot and compare
Why: Same vertex at the origin, but the curve falls away instead of rising.
Figure (svg): The solution to Worked example wider and opening downward shown as a ladder of expressions, one row per algebraic move
\[ y = -4x^2 \quad \text{narrower, opening down} \]
Verify: check the range
Why: Every output is negative four times a non-negative number, so y is at most zero with equality only at the vertex. The range is the mirror image of the parent's, which is what a downward-opening parabola must have.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.1 Graph Quadratic Functions in Standard Form §4.1, pp. 237-237
Trap
\[ y = 4x^2 \]
Say the parabola is wider because the coefficient is larger
Why: Larger is read as bigger in every sense, including horizontally.
But every height is four times the parent's, so the curve reaches any given height at a QUARTER of the horizontal distance.
\[ y = 4x^2 \]
Judge width by what happens to the OUTPUT
Why: Multiplying by 4 quadruples every height, which is a vertical stretch. The narrowness is a consequence of that.
\[ \text{at } y = 4: \; x^2 \text{ gives } x = \pm 2, \; 4x^2 \text{ gives } x = \pm 1 \]
The stretched parabola reaches height 4 at x equal to plus or minus 1 rather than plus or minus 2 — half the width, exactly as a stretch predicts.
Tweak it
One dial, two visible effects.
Parameter explorer
What changes as a varies, and what stays fixed?
\[ y = {a}x^2 \]
Prediction
Commit before reasoning.
Predict first
Which of y equals 3x squared, y equals negative 5x squared, and y equals one half x squared is narrowest?
Correct: y equals negative 5x squared — the size of its coefficient is largest.
\[ \lvert -5 \rvert > \lvert 3 \rvert > \lvert \tfrac{1}{2} \rvert \]
Why: Width depends only on the SIZE of a, and five exceeds three which exceeds one half. The negative sign flips the parabola downward but does not affect how narrow it is. Judging width by the signed value rather than the size is the commonest error here, and it makes the negative case look small when it is in fact the most stretched.
Comparison
Fill the blanks. The coefficient a behaves identically in both families.
Comparison matrix
| Feature | y = a|x| (Lesson 2.7) | y = ax^2 (this lesson) |
|---|---|---|
| Shape | a V with two straight branches | a smooth parabola |
| |a| > 1 | narrower | narrower |
| a < 0 | opens downward | opens downward |
| Vertex when b and c are 0 | the origin | the origin |
Only the first row differs. Everything you learned about a in Lesson 2.7 transfers directly, which is what makes a new function family quick to learn once you have the habit.
Section
Section 3
Concept
In y equals a x squared plus c, the constant c slides the parabola vertically. When b is zero the vertex stays on the vertical axis, at the point (0, c).
\[ y = -\tfrac{1}{2}x^2 + 3 \]
The y-intercept of any quadratic in standard form is c, because substituting zero for x removes both the other terms.
Figure (svg): A parabola shifted three units up, with the parent shown for comparison
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.1 Graph Quadratic Functions in Standard Form §4.1, pp. 237-237 — Graph a function of the form y = ax^2 + c
Picture it
Example 2: y equals negative one half x squared plus 3.
Figure (svg): A parabola shifted three units up, with the parent shown for comparison
Three changes from the parent, and each comes from a different feature: the size of a widens it, the sign of a flips it, and c lifts it.
Worked example
Example 2, with the comparison spelled out.
\[ \text{Graph } y = -\tfrac{1}{2}x^2 + 3 \text{ and compare it with } y = x^2. \]
Choose inputs that are multiples of 2
Why: The coefficient is a half, so even inputs keep the outputs whole.
\[ x = -4, -2, 0, 2, 4 \]
Compute the outputs
Why: At plus or minus four: half of sixteen is eight, negated is negative eight, plus three is negative five. At plus or minus two: negative two plus three is one. At zero: three.
\[ -5, 1, 3, 1, -5 \]
Plot and join smoothly
Why: The vertex is at (0, 3) and the curve falls away on both sides.
\[ \text{vertex } (0, 3) \]
Compare with the parent
Why: Same axis of symmetry; opens down instead of up; wider; vertex three units higher.
Figure (svg): The solution to Worked example all three effects at once shown as a ladder of expressions, one row per algebraic move
\[ y = -\tfrac{1}{2}x^2 + 3, \quad \text{vertex } (0, 3) \]
Verify: confirm the vertex is the maximum
Why: Because a is negative, the vertex is the highest point, so 3 should be the largest output. Testing x equal to 1 gives negative one half plus three, which is 2.5 — below 3, as required. Every other input gives less still.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.1 Graph Quadratic Functions in Standard Form §4.1, pp. 237-237
Matching
When b is zero the vertex is on the vertical axis.
Match the pairs
Why: In every case the first coordinate is zero, because with no x term the squared quantity is smallest at x equal to zero. The second coordinate is just c, which is also the y-intercept. When b is not zero this all changes, which is the subject of the next section.
Worked example
Guided Practice 2. The same idea with a negative constant.
\[ \text{Graph } y = -x^2 - 5 \text{ and compare it with } y = x^2. \]
Read the coefficient
Why: Negative one: size one, so the same width as the parent; negative, so it opens downward.
Read the constant
Why: Negative five, so the vertex sits five units below the origin.
\[ \text{vertex } (0, -5) \]
Tabulate a few values
Why: At plus or minus one, negative one minus five is negative six; at plus or minus two, negative four minus five is negative nine.
\[ (1, -6), (2, -9) \]
Compare with the parent
Why: Same width, flipped, and slid down five.
Figure (svg): The solution to Worked example a downward shift shown as a ladder of expressions, one row per algebraic move
\[ y = -x^2 - 5, \quad \text{vertex } (0, -5) \]
Verify: check the maximum value
Why: The vertex is the highest point since a is negative, so no output should exceed negative five. At x equal to 3 the value is negative fourteen, comfortably below. The whole graph lies at or below y equal to negative five.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.1 Graph Quadratic Functions in Standard Form §4.1, pp. 237-237
Trap
\[ y = x^2 + 3 \]
Slide the parabola three units to the right
Why: The constant is treated like the h of Lesson 2.7's absolute value form.
But h lived INSIDE the brackets there. A constant added outside moves the graph vertically, not horizontally.
\[ y = x^2 + 3, \quad \text{vertex } (0, 3) \]
Substitute x equal to zero to locate the vertex
Why: The constant is added after the squaring, so it raises every output by 3 — including the one at the vertex.
\[ y(0) = 0 + 3 = 3 \]
Compare y equals the quantity x minus 3, squared, which DOES shift right by three. Where the number sits relative to the squaring decides everything, exactly as it did in Lesson 2.7.
Fill the middle
Example 2's function at x equal to negative four.
Fill in the blanks
y = -\tfrac-5___(-4)^2 + 3 = -\tfrac______(16) + 3 = ___
Why: Negative four squared is positive sixteen, half of which is eight, and the negative sign makes it negative eight; adding three gives negative five. The brackets around the substituted negative are what keep the square positive — the same discipline as Lesson 1.2.
Discrimination
Three separate effects, three separate causes.
Sort into buckets
Sort each described change by what causes it.
Prediction
Commit before computing.
Predict first
What is the y-intercept of y equals 2x squared minus 8x plus 6?
Correct: 6 — the constant term.
\[ y(0) = 2(0)^2 - 8(0) + 6 = 6 \]
Why: Substituting x equal to zero makes both the x squared and the x terms vanish, leaving the constant. So the y-intercept of any quadratic in standard form is c, whatever a and b are. This holds even when b is not zero, which is why the intercept is a reliable free point when graphing the general case — and its reflection across the axis of symmetry gives a second one.
Section
Section 4
Concept
When b is not zero the vertex leaves the vertical axis. Its x-coordinate is negative b over 2a, and substituting that back gives the y-coordinate. The axis of symmetry is the vertical line through it.
\[ x = -\frac{b}{2a} \]
Once the vertex is plotted, the y-intercept c and its reflection across the axis give two more points for free.
Figure (svg): The parabola y equals 2x squared minus 8x plus 6 with its vertex, axis of symmetry, intercept and reflected points marked
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.1 Graph Quadratic Functions in Standard Form §4.1, pp. 237-238 — Properties of the Graph of y = ax^2 + bx + c
Picture it
Example 3: y equals 2x squared minus 8x plus 6.
Figure (svg): The parabola y equals 2x squared minus 8x plus 6 with its vertex, axis of symmetry, intercept and reflected points marked
The vertex, the intercept, its reflection, and one extra point with its reflection give five points — more than enough to draw the curve confidently.
Worked example
Example 3, all six steps.
\[ \text{Graph } y = 2x^2 - 8x + 6. \]
Identify the coefficients and the direction
Why: a is 2, b is negative 8, c is 6. Since a is positive the parabola opens up.
\[ a = 2, b = -8, c = 6 \]
Compute the x-coordinate of the vertex
Why: Negative b over 2a is 8 over 4.
\[ x = 2 \]
Compute the y-coordinate
Why: Two times four minus sixteen plus six.
\[ y = -2 \]
Plot the vertex, the axis, and the y-intercept
Why: The vertex is (2,-2), the axis is x equal to 2, and the intercept is (0,6); reflecting it gives (4,6).
\[ (2, -2), (0, 6), (4, 6) \]
Add one more point and its reflection
Why: At x equal to 1 the value is 2 minus 8 plus 6, which is 0; reflecting gives (3,0).
\[ (1, 0)\text{ and } (3, 0) \]
Figure (svg): The solution to Worked example graph a general quadratic shown as a ladder of expressions, one row per algebraic move
\[ y = 2x^2 - 8x + 6, \quad \text{vertex } (2, -2) \]
Verify: check the two reflected points satisfy the equation
Why: At x equal to 4: 2 times 16 minus 32 plus 6 is 32 minus 32 plus 6, which is 6 — matching the reflected intercept. At x equal to 3: 18 minus 24 plus 6 is 0 — matching the other reflection. The symmetry gave two correct points without computing them, and the check confirms it.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.1 Graph Quadratic Functions in Standard Form §4.1, pp. 238-238
Fill the middle
Guided Practice 4: y equals x squared minus 2x minus 1.
Fill in the blanks
x = -\frac1___ = -\frac______ = ___
Why: Negative b is negative of negative two, which is positive two, and 2a is 2, so the x-coordinate is 1. Substituting gives 1 minus 2 minus 1, which is negative 2, so the vertex is (1, -2). The double negative is where this formula most often goes wrong.
Worked example
Guided Practice 5. The formula handles it without difficulty.
\[ \text{Find the vertex and axis of symmetry of } y = 2x^2 + 6x + 3. \]
Identify the coefficients
Why: a is 2, b is 6, c is 3.
\[ a = 2, b = 6, c = 3 \]
Compute the x-coordinate
Why: Negative six over four, which reduces to negative three halves.
\[ x = -1.5 \]
Substitute to get the y-coordinate
Why: Two times 2.25 is 4.5; six times negative 1.5 is negative 9; plus 3.
\[ 4.5 - 9 + 3 = -1.5 \]
State the vertex and axis
Why: The vertex is at negative 1.5, negative 1.5 and the axis is the vertical line there.
\[ \text{vertex } (-1.5, -1.5) \]
Figure (svg): The solution to Worked example a vertex with fractional coordinates shown as a ladder of expressions, one row per algebraic move
\[ \text{vertex } \left(-\tfrac{3}{2}, -\tfrac{3}{2}\right) \]
Verify: check symmetry using the y-intercept
Why: The intercept is (0, 3), which is 1.5 to the right of the axis, so its reflection is 1.5 to the left, at (-3, 3). Substituting: 2 times 9 minus 18 plus 3 is 18 minus 18 plus 3, which is 3 — confirming both the reflection and the axis.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.1 Graph Quadratic Functions in Standard Form §4.1, pp. 238-238
Error analysis
A student computes the vertex of y equals 2x squared minus 8x plus 6.
Annotate
On: \( x = \frac{b}{2a} = \frac{-8}{4} = -2 \)
After computing a vertex, evaluate the function one unit either side. For an upward parabola both must be higher than the vertex value, and for a downward one both must be lower.
Ranking
Graphing a quadratic in standard form.
Put in order
Why: The vertex must exist before the axis of symmetry can be drawn, and the axis must exist before anything can be reflected across it. Reflecting the intercept is free information that most students skip, and it is what makes five points available for the price of three calculations.
Prediction
Commit before computing.
Predict first
For y equals 3x squared minus 18x plus 20, where is the axis of symmetry?
Correct: x = 3.
\[ x = -\frac{-18}{2(3)} = \frac{18}{6} = 3 \]
Why: Negative b over 2a is negative of negative 18, over 6, which is 18 over 6, or 3. The axis of symmetry is the vertical line through the vertex, so its equation is x equals 3. Note that it is an equation of a line, not a number — writing the answer as 3 alone would name the x-coordinate rather than the axis.
Explain it to yourself
The vertex formula looks arbitrary until you connect it to the symmetry.
\[ x = -\frac{b}{2a} \]
Discussion prompt
The y-intercept is at x equal to zero. Find the OTHER x value that gives the same output c, and explain why the axis of symmetry must sit halfway between them.
Hint: Set a x squared plus b x plus c equal to c and solve.
Answer:
\[ ax^2 + bx + c = c \;\Longrightarrow\; x(ax + b) = 0 \;\Longrightarrow\; x = 0 \text{ or } x = -\tfrac{b}{a} \]
So the graph passes through the same height c at x equal to zero and at x equal to negative b over a. The axis of symmetry must be midway between any two points at equal height.
The midpoint of 0 and negative b over a is negative b over 2a — which is the formula. It is not arbitrary at all; it is the midpoint of a pair the symmetry hands you.
Section
Section 5
Concept
For a parabola opening up, the vertex's y-coordinate is the minimum value of the function; for one opening down, it is the maximum. That single fact is what makes quadratics useful for optimisation.
\[ a > 0 \;\Rightarrow\; \text{minimum} \qquad a < 0 \;\Rightarrow\; \text{maximum} \]
The minimum or maximum is the y-coordinate, not the x-coordinate. The x-coordinate says where it happens; the y-coordinate says what it is.
Figure (svg): Two columns contrasting a parabola with a minimum against one with a maximum
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.1 Graph Quadratic Functions in Standard Form §4.1, pp. 238-239 — Minimum and Maximum Values
Picture it
The sign of a decides.
Figure (svg): Two columns contrasting a parabola with a minimum against one with a maximum
A parabola opening up has a minimum and no maximum — it rises without bound on both sides. The reverse holds for one opening down.
Worked example
Example 4. Decide which kind first, then compute.
\[ \text{Does } y = 3x^2 - 18x + 20 \text{ have a minimum or a maximum? Find it.} \]
Read the sign of a
Why: Three is positive, so the parabola opens up and the vertex is the lowest point.
\[ a > 0: a\text{ minimum} \]
Find the x-coordinate of the vertex
Why: Negative of negative eighteen, over six.
\[ x = 3 \]
Substitute to find the value
Why: Three times nine is 27, minus 54, plus 20.
\[ 27 - 54 + 20 = -7 \]
State the answer correctly
Why: The minimum VALUE is negative 7, and it occurs at x equal to 3.
\[ \text{minimum is } -7 \]
Figure (svg): The solution to Worked example find a minimum value shown as a ladder of expressions, one row per algebraic move
\[ \text{minimum} = -7 \text{ at } x = 3 \]
Verify: evaluate either side of the vertex
Why: At x equal to 2 the value is 12 minus 36 plus 20, which is negative 4. At x equal to 4 it is 48 minus 72 plus 20, also negative 4. Both are above negative 7, and they are equal to each other — which confirms both the minimum and the symmetry about x equal to 3.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.1 Graph Quadratic Functions in Standard Form §4.1, pp. 239-239
Sorting
Read the sign of the leading coefficient.
Sort into buckets
Sort each function by whether its vertex is a minimum or a maximum.
Only the sign of a matters. The other coefficients affect where the extreme is and what it equals, never which kind it is.
Worked example
Guided Practice 7.
\[ \text{Find the minimum value of } y = 4x^2 + 16x - 3. \]
Confirm it is a minimum
Why: a is 4, which is positive, so the parabola opens up.
Find the x-coordinate
Why: Negative sixteen over eight.
\[ x = -2 \]
Substitute
Why: Four times four is 16, plus 16 times negative 2 which is negative 32, minus 3.
\[ 16 - 32 - 3 = -19 \]
State the value
Why: The minimum value is negative 19.
\[ \text{minimum is } -19 \]
Figure (svg): The solution to Worked example another minimum shown as a ladder of expressions, one row per algebraic move
\[ \text{minimum} = -19 \text{ at } x = -2 \]
Verify: test a nearby input
Why: At x equal to negative 1 the value is 4 minus 16 minus 3, which is negative 15 — above negative 19 as required. At x equal to negative 3 it is 36 minus 48 minus 3, also negative 15, matching by symmetry.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.1 Graph Quadratic Functions in Standard Form §4.1, pp. 239-239
Error analysis
A student answers Example 4 with the wrong coordinate.
Annotate
On: \( x = 3 \;\Longrightarrow\; \text{the minimum value is } 3 \)
Say the answer as a sentence: the minimum value is -7 and it occurs at x = 3. A sentence makes the two roles impossible to confuse.
Real world
Example 5: a go-cart track has 380 racers a week at 35 dollars each, and gains 20 more racers for every dollar the price is cut.
Discussion prompt
Build the revenue function, find the price reduction that maximises it, and state the maximum revenue. Explain why the model has a maximum rather than a minimum.
Hint: Revenue is price times attendance, and both change with the reduction.
Answer:
\[ R(x) = (35 - x)(380 + 20x) = -20x^2 + 320x + 13\,300 \]
\[ x = -\frac{320}{2(-20)} = 8, \qquad R(8) = 14\,580 \]
Reducing the price by 8 dollars gives a maximum weekly revenue of 14,580 dollars. The model has a maximum because a is negative twenty — cutting the price raises attendance but lowers the price per racer, and past a point the second effect dominates.
Fill the middle
Example 5, step two.
Fill in the blanks
(35 - x)(380 + 20x) = 13300 + 700x - 380x - 20x^2 = -20x^2 + 320x + 13300
Why: Seven hundred x minus 380x is 320x, and writing the terms in descending powers gives negative 20x squared plus 320x plus 13,300. Expanding into standard form is what makes the vertex formula usable — the factored form is easier to build from the situation but harder to optimise.
Socratic
One question, and nothing else on this slide.
\[ R(x) = -20x^2 + 320x + 13\,300 \]
Discussion prompt
The vertex is at (8, 14580). What does each coordinate mean to the track owner, and what would you tell them about reducing the price by 12 dollars instead? Why does the model eventually predict falling revenue?
Hint: Evaluate R at 12 and compare.
Answer:
The 8 is the price reduction in dollars and the 14,580 is the resulting weekly revenue. They are different quantities in different units, which is why reporting one for the other would be meaningless.
\[ R(12) = -20(144) + 3840 + 13\,300 = 14\,260 \]
A 12-dollar cut earns 320 dollars less than an 8-dollar cut. Past the vertex, each further reduction brings in extra racers but the price per racer has fallen so far that total revenue declines — which is exactly what a downward parabola describes.
Comparison
Fill the blanks. Each coefficient answers one question.
Comparison matrix
| Feature | Where it comes from | What it tells you |
|---|---|---|
| Direction of opening | the sign of a | up if positive, down if negative |
| Width | the size of a | narrower if bigger than 1 |
| x-coordinate of the vertex | -b/(2a) | where the extreme value occurs |
| y-intercept | the constant c | one free point, plus its reflection |
| Minimum or maximum value | the vertex's y-coordinate | the extreme value itself |
Three of the five come from a and b alone. The constant contributes the intercept and, when b is zero, the vertex height as well.
Pattern
One routine graphs any quadratic in standard form.
Step four is the one most often skipped, and it is the cheapest information in the lesson — two points for no extra calculation.
OpenStax Algebra and Trigonometry 2e, §5.1 Quadratic Functions §5.1
Check
The vertex formula. Watch the sign.
Check your understanding
What is the vertex of y = 2x^2 - 8x + 6?
Answer: A
Why: The x-coordinate is -b/(2a), which is 8/4, or 2. Substituting gives 8 - 16 + 6, which is -2. Since a is positive, this is the lowest point.
Check
Minimum or maximum, and which coordinate.
Check your understanding
Find the minimum value of y = 4x^2 + 16x - 3.
Answer: A
Why: Since a is positive there is a minimum. The vertex is at x = -16/8 = -2, and substituting gives 16 - 32 - 3, which is -19.
Check
Comparing with the parent. Size and sign separately.
Check your understanding
How does the graph of y = -(1/2)x^2 + 3 compare with the graph of y = x^2?
Answer: A
Why: The size of a is one half, which is less than 1, so the parabola is wider. The sign is negative, so it opens downward. The constant 3 raises the vertex to (0, 3).
Real world
A shop sells 200 items a week at 40 dollars each and estimates that each 1 dollar price rise loses 5 sales a week.
Discussion prompt
Build the revenue function, find the price that maximises revenue, and say what the maximum is. Then say why raising the price indefinitely does not keep raising revenue.
Hint: Let x be the price rise in dollars, and write both the price and the quantity in terms of it.
Answer:
\[ R(x) = (40 + x)(200 - 5x) = -5x^2 + 200x + 8000 \]
\[ x = -\frac{200}{2(-5)} = 20, \qquad R(20) = 10\,000 \]
A 20-dollar rise, to 60 dollars, gives a maximum weekly revenue of 10,000 dollars from 100 sales. Beyond that, each further rise loses more in volume than it gains in price.
The model also has a hard limit the algebra reveals: at x equal to 40 the quantity 200 minus 5x reaches zero, so nobody buys anything and revenue is zero. The parabola's downward opening encodes that squeeze from both ends.
Commit first
Answer, then rate your confidence honestly.
Predict first
Does every quadratic function have both a minimum and a maximum value?
Correct: No — a quadratic has exactly one of the two, decided by the sign of a.
Restricting the domain changes this: on a closed interval a quadratic does have both, and the maximum then sits at an endpoint. That is a genuinely different question and one worth keeping separate.
Why: A parabola opening upward rises without bound on both sides, so it has a lowest point and no highest one. Opening downward, the reverse. The vertex is always one extreme and there is never another, which is precisely why quadratics are the natural tool for optimisation problems with a single best answer.
Explain it
They can graph lines and have never met a curve given by an equation.
Discussion prompt
In four sentences or fewer, explain what a parabola is, how to find its vertex from standard form, and the shortcut that gives two extra points for free.
Hint: The shortcut involves the y-intercept and the symmetry.
Answer:
A parabola is the U-shaped graph of any function with an x squared term, and its turning point is the vertex. From y equals a x squared plus b x plus c, the vertex sits at x equal to negative b over 2a, and substituting that value back gives its height.
The free shortcut: the graph always passes through (0, c), and because it is symmetric about the vertical line through the vertex, reflecting that point across the line gives a second one at no extra cost.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the sign, write the formula as negative b over 2a and substitute b with its own sign in brackets. For min against max, read only the sign of a. For value against location, state the answer as a full sentence. For revenue models, write price times quantity with both expressed in terms of the same variable. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
In the middle of a page draw one coordinate plane and graph the parent parabola in pencil, marking its vertex and drawing its axis of symmetry as a dashed line. In a second colour add a narrower parabola and in a third a wider one that opens downward, labelling each with its equation and writing three words about what changed. Below, take a quadratic of your own in standard form with b not zero, compute the vertex using the formula with every sign written out, plot it, plot the y-intercept, reflect that intercept across the axis, add one more point with its reflection, and draw the curve through all five. In the margin write whether your parabola has a minimum or a maximum and what that value is, as a full sentence. Finally, invent a small revenue situation, write the function as price times quantity, expand it, and find the vertex.
If your margin sentence names only a number, rewrite it to say both what the extreme value is and where it occurs. Those are two different coordinates and word problems ask for different ones.
Recap
Five things, and the fourth is the formula the whole chapter uses.
| If you see | Then |
|---|---|
| An x squared term | The graph is a parabola, not a line |
| A negative leading coefficient | It opens down; the vertex is a maximum |
| A coefficient bigger than 1 in size | Narrower than the parent |
| Standard form with b not zero | Vertex at -b/(2a) |
| A revenue model | Expand to standard form, then find the vertex |
Lesson 4.2 rewrites the same parabolas in two other forms — vertex form and intercept form — each of which makes something visible that standard form hides.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.1 Graph Quadratic Functions in Standard Form §4.1, pp. 236-241 — everything on these slides traces back here
Want this taught 1-on-1? Alexander tutors Algebra 2 — $55/session, free consultation.