UNIT 3Quadratic functionsLesson 3 of 31 Solving Quadratics by Factoring 2 Solving Quadratics by Square Roots3 Graphing Quadratic Functions

Graphing Quadratic Functions

A1.7Algebra 1Functions20 practice questions

The graph of y = ax² + bx + c is a U-shaped curve called a parabola. A few key points tell you its whole shape.

What you'll be able to do

Tell which way a parabola opens, and if it has a maximum or a minimum.
Find the axis of symmetry and the vertex.
Find the intercepts, and give the domain and range.
Solve launch and fence stories with the vertex.

The big idea

Up or down. If a > 0, the parabola opens up and its lowest point is a minimum. If a < 0, it opens down and its highest point is a maximum.

The axis of symmetry. It is the line x = −b/(2a). The parabola is the same on both sides of it.

The vertex. Put x = −b/(2a) into the function to get the y-value. That point is the vertex.

Intercepts. The y-intercept is (0, c). The x-intercepts are where y = 0, so solve ax² + bx + c = 0.

Mirror points. Points with the same y-value sit the same distance from the axis, one on each side.

Domain and range. The domain is all real numbers. The range starts or ends at the vertex's y-value.

Words you need

Parabola: The U-shaped graph of a quadratic function.
Quadratic function: A function y = ax² + bx + c, where a is not 0.
Vertex: The turning point of a parabola. It is the highest or lowest point.
Axis of symmetry: The vertical line through the vertex. The parabola is the same on both sides.
Maximum: The greatest y-value. A parabola that opens down has one.
Minimum: The least y-value. A parabola that opens up has one.
Intercept: A point where the graph crosses an axis.
Domain: All the x-values you can put into a function.
Range: All the y-values a function gives out.
Standard form: y = ax² + bx + c, with the terms in order of their exponents.
Coefficient: The number in front of a letter. In −16t², it is −16.
Leading coefficient: The coefficient of the term with the highest exponent. It is a.
Constant: A term with no letter, like c.
Factor: A part you multiply. To factor means to write something as a product.
Product: The answer when you multiply.
Integer: A whole number or its negative, like −3, 0 or 7.
Real number: Any number on the number line.
Exponent: The small raised number. In x², it is 2.
Power: A base with an exponent, like x².
Base: The bottom of a shape, like the base of a pier. In a power like 3², the base is 3.
Variable: A letter that stands for a number, like x or t.
Expression: Numbers and letters joined by +, −, × or ÷. It has no = sign.
Simplify: Write it in a shorter form with the same value.
Zero: An x-value that makes a function equal 0. It is an x-intercept.
Square root: A number that gives the first number when you multiply it by itself.
Denominator: The bottom number of a fraction.

Reading a parabola from its equation

Feature	How to find it
Opens up or down	Up if a > 0, down if a < 0
Axis of symmetry	x = −b/(2a)
Vertex	Put the axis x-value into the function
y-intercept	(0, c)
x-intercepts	Solve ax² + bx + c = 0

Feature

How to find it

Opens up or down

Up if a > 0, down if a < 0

Axis of symmetry

x = −b/(2a)

Vertex

Put the axis x-value into the function

y-intercept

(0, c)

x-intercepts

Solve ax² + bx + c = 0

Worked examples

Example 1

For y = 2x² − 8x + 3, find the axis of symmetry, the vertex and the y-intercept. Is the vertex a maximum or a minimum?

a = 2 and b = −8. The axis is x = −(−8)/(2 · 2) = 2.
At x = 2, y = 2(4) − 8(2) + 3 = −5. The vertex is (2, −5).
a = 2 is positive, so the parabola opens up. The vertex is a minimum.
The y-intercept is (0, 3).

Answer: Axis x = 2, vertex (2, −5), a minimum, y-intercept (0, 3)

Example 2

Find the x-intercepts of y = x² − 2x − 8. Then use them to find the vertex.

Set y = 0: x² − 2x − 8 = 0, so (x − 4)(x + 2) = 0.
The x-intercepts are (4, 0) and (−2, 0).
The axis is halfway between them: x = (4 + (−2)) ÷ 2 = 1.
At x = 1, y = 1 − 2 − 8 = −9. The vertex is (1, −9).

Answer: x-intercepts (−2, 0) and (4, 0), vertex (1, −9)

Example 3

A ball is thrown upward. Its height in feet after t seconds is h = −16t² + 40t + 3. When is it highest, and how high does it go?

a = −16 is negative, so the vertex is a maximum.
The axis is t = −40/(2 · (−16)) = 1.25.
At t = 1.25, h = −16(1.5625) + 40(1.25) + 3 = −25 + 50 + 3 = 28.
The ball is highest after 1.25 seconds, at 28 feet.

Answer: After 1.25 seconds, at 28 feet

Example 4

Ama has 40 meters of fence for a rectangle. What width gives the largest area?

Two widths and two lengths use 40 meters, so a width plus a length is 20. If the width is x, the length is 20 − x.
The area is A = x(20 − x) = −x² + 20x.
The vertex is at x = −20/(2 · (−1)) = 10.
A width of 10 meters gives the largest area: 10 × 10 = 100 square meters.

Answer: A width of 10 meters, for 100 square meters

Common mistakes

Dropping the minus sign. For y = x² − 6x + 1, the axis is x = −(−6)/2 = 3, not −3.

Mixing up maximum and minimum. If a < 0, the parabola opens down. The vertex is a maximum.

Giving the range as all real numbers. A parabola stops at its vertex. If the vertex is (2, 9) and it opens down, the range is y ≤ 9.

Reading the wrong coordinate. The maximum value is the vertex's y-value, not its x-value.

Practice: 20 questions

11 of 20 are word problems. For a word problem, sketch it first and mark what you know.

Tier 1 Foundations Use the rule directly

Write the quadratic function y = 14 − 9x − 5x² in standard form. Then state whether its parabola opens up or down, whether its vertex represents a maximum or a minimum, and what its y-intercept is. Give the y-intercept as an ordered pair (x, y).
Jiro launches a water rocket straight up from a rooftop. The graph shows the rocket's height y, in feet, above the ground x seconds after launch. From the graph, find the starting height of the rocket, the maximum height and when it happens, and how many seconds the rocket is in the air before hitting the ground. Give heights in feet and times in seconds. Word problem
Fatima throws a ball upward from a platform. Its height y, in feet, after x seconds follows a parabola whose axis of symmetry is the vertical line x = 3. At x = 1 second, the ball is at a height of 84 feet, which is the point (1, 84) on the graph. Find the mirror point of (1, 84) across the axis of symmetry, and state the time when the ball is at a height of 84 feet again on its way down. Give the mirror point as an ordered pair (x, y) and the time in seconds. Word problem
Nkechi wants to find the axis of symmetry and vertex of the parabola y = x² − 8x + 15. She writes x = −8/(2 · 1) = −4, substitutes x = −4 to get y = (−4)² − 8(−4) + 15 = 63, and concludes that the vertex is (−4, 63). Identify the sign mistake in Nkechi's first step, and find the correct axis of symmetry and vertex. Give the axis of symmetry as an equation and the vertex as an ordered pair (x, y).
Compare the two parabolas y = −x² + 12 and y = −2x² + 9. Which parabola has the higher maximum value, and by how many units? Give the maximum value of each parabola and the difference as integers.

Tier 2 Intermediate One more step to think through

Laszlo has 24 meters of fencing to enclose a four-sided rectangular herb garden of width x meters and length (12 − x) meters. The area y, in square meters, of the garden is given by y = −x² + 12x. Find the axis of symmetry and the vertex of this parabola, and state what the coordinates of the vertex mean for Laszlo's garden. Give the axis of symmetry as an equation and the vertex as an ordered pair (x, y). Word problem
Ximena throws a stone straight up from a bridge 32 feet above a river with an initial upward velocity of 56 feet per second. The stone's height y, in feet, above the water after x seconds is modeled by y = −16x² + 56x + 32. Find the y-intercept and both x-intercepts of the graph, and explain what the y-intercept and the positive x-intercept mean for the stone. Give all intercepts as ordered pairs (x, y). Word problem
Darius launches a signal flare straight up from a 20-foot platform with an initial upward velocity of 32 feet per second, so its height y, in feet, after x seconds is modeled by the quadratic function y = −16x² + 32x + 20. Find the vertex of the parabola, and state the domain and range of the function y = −16x² + 32x + 20 over all real numbers. Give the vertex as an ordered pair (x, y), the domain in words, and the range as an inequality. Word problem
Mireille finds the vertex of the parabola y = −2x² + 12x − 11 correctly as (3, 7). She then states that the parabola has a minimum value of 7 and that its range is all real numbers. Identify both mistakes in Mireille's statement, and give the correct description of the vertex value and the correct range as an inequality.
Compare the x-intercepts of two parabolas. Parabola A has the equation y = x² − 4x − 21. Parabola B has the equation y = x² − 10x + 9. Which parabola has the wider horizontal spread between its two x-intercepts, and by how many units? Give the x-intercepts and the spread for each parabola as integers.

Tier 3 Advanced Several steps - find the hidden structure

Bram launches a model rocket upward from a cliff 72 feet above a canyon floor with an initial upward velocity of 56 feet per second. Its height h, in feet, above the canyon floor after t seconds is given by the formula h = −16t² + 56t + 72. Find when the rocket reaches its maximum height, what that maximum height is, and when the rocket hits the canyon floor. Give times as decimals or fractions in simplest form in seconds, and height in feet. Word problem
Kavya is building a rectangular sheep pen along a straight stone wall that is already in place. The stone wall forms one long side of the pen, so she needs fencing only for the two widths of x meters each and the one side parallel to the wall. She has 80 meters of fencing in total. Write an equation in standard form for the area A, in square meters, of the pen in terms of x, and find the width and length that give the largest possible area, along with that maximum area. Word problem
Teodoro designs a parabolic steel arch to support a footbridge over a creek. In his coordinate plan, x is the horizontal distance in feet from the left bank, and y is the height of the arch in feet above the water surface, where negative y-values represent concrete footings below the water surface. The graph shows the arch for x from 0 to 10. From the graph, find both x-intercepts, the horizontal width of the arch at water level where y = 0, and the maximum height of the arch above the water. Give the x-intercepts as ordered pairs (x, y) and the width and height in feet. Word problem
Amina wants to find the vertex and range of y = −3x² − 18x − 20. She calculates x = −(−18)/(2 · −3) = −3. Next, she substitutes x = −3 by writing y = (−3 · −3)² − 18(−3) − 20 = 81 + 54 − 20 = 115, and states that the range is y ≤ 115. Explain the order-of-operations mistake Amina made when finding y, and find the correct vertex and range. Give the vertex as an ordered pair (x, y) and the range as an inequality.
Two flare guns are fired straight up at the same time. Flare A's height h, in feet, after t seconds is given by h = −16t² + 64t + 15. Flare B's height h, in feet, after t seconds is given by h = −16t² + 48t + 40. Which flare reaches the higher maximum height, and by how many feet? Give the maximum height of each flare and the difference in feet. Word problem

Tier 4 Challenge Stretch problems, some above grade level

Stellan has 120 feet of wire fencing to build a rectangular dog kennel that is divided into three side-by-side pens by two interior fences running parallel to the outer width of x feet. All four parallel width segments have length x feet, and the two long outer sides perpendicular to them each have length L feet. Write an equation in standard form for the total enclosed area A, in square feet, of the kennel in terms of x, and find the width x, the total length L, and the maximum total area. Word problem
Femi launches a projectile straight up from the ground at t = 0 seconds, so its height h, in feet, after t seconds follows h = −16t² + vt. On its way up, the projectile passes a height of 96 feet at t = 2 seconds. On its way down, it passes the same 96-foot height at t = 3 seconds. Find the time when the projectile reaches its maximum height, the time when it lands back on the ground, the initial upward velocity v in feet per second, and the maximum height in feet. Word problem
Giselle wants to write a quadratic function in standard form y = ax² + bx + c whose graph has x-intercepts at (1, 0) and (5, 0) and a maximum value of 12. She starts with y = a(x − 1)(x − 5), finds the axis of symmetry x = 3, substitutes (3, 12) to get 12 = a(3 − 1)(3 − 5) = −4a, and then writes a = 3 to get y = 3x² − 18x + 15. Identify the sign mistake Giselle made when solving for a, explain how you can tell immediately from the word "maximum" that y = 3x² − 18x + 15 cannot be right, and write the correct quadratic function in standard form.
Let k be a positive integer. Parabola A has the equation y = −x² + 4kx. Parabola B has the equation y = −4x² + 12kx. Which parabola has the higher maximum value, and which parabola has the wider spread between its x-intercepts? Express the maximum value and the x-intercept spread of each parabola in terms of k.
A cubic polynomial function has degree 3. Its zeros, or x-intercepts, are the x-values that make y = 0, and those zeros divide the x-axis into intervals where the graph lies either entirely above the x-axis with y > 0 or entirely below the x-axis with y < 0. Consider the cubic function y = x³ − 2x² − 8x. Factor the right side completely to find all three x-intercepts in order from least to greatest, and determine whether the graph is above or below the x-axis on the interval between the first two x-intercepts and on the interval between the last two x-intercepts. Above grade level

Answers

Tier 1 Foundations

Standard form: y = −5x² − 9x + 14; the parabola opens down, has a maximum, and has a y-intercept of (0, 14). Rearrange the terms in descending order of exponents to get y = −5x² − 9x + 14, where a = −5, b = −9, and c = 14. Because the leading coefficient a = −5 is negative, the parabola opens down and its vertex is the highest point, giving a maximum. Substituting x = 0 gives y = 14, so the y-intercept is (0, 14).
Starting height: 80 feet; maximum height: 144 feet at 2 seconds; time in the air before hitting the ground: 5 seconds. Read the y-intercept (0, 80) on the vertical axis to find the starting height of 80 feet. Locate the vertex at the peak of the parabola, (2, 144), which shows a maximum height of 144 feet at 2 seconds. Read the positive x-intercept (5, 0) on the horizontal axis to see that the rocket hits the ground after 5 seconds.
Mirror point: (5, 84); the ball is at 84 feet again at 5 seconds. The point (1, 84) has x-coordinate 1, which is 3 − 1 = 2 units to the left of the axis of symmetry x = 3. Its mirror point lies 2 units to the right of x = 3 at x = 3 + 2 = 5 with the same y-coordinate, giving (5, 84). Thus, the ball passes a height of 84 feet again at 5 seconds.
Nkechi wrote b/(2a) instead of −b/(2a) and dropped the negative sign in front of b; the correct axis of symmetry is x = 4 and the vertex is (4, −1). In y = x² − 8x + 15, a = 1 and b = −8, so the axis of symmetry is x = −(−8)/(2 · 1) = 8/2 = 4, not −4. Substitute x = 4 into the equation to get y = 4² − 8(4) + 15 = 16 − 32 + 15 = −1. The vertex is (4, −1).
The parabola y = −x² + 12 has the higher maximum value by 3 units; its maximum value is 12, while the maximum value of y = −2x² + 9 is 9. Since b = 0 in both equations, the axis of symmetry for each parabola is x = 0. Substituting x = 0 into y = −x² + 12 gives a maximum value of 12, and substituting x = 0 into y = −2x² + 9 gives a maximum value of 9. Comparing the two maxima gives 12 − 9 = 3 units.

Tier 2 Intermediate

Axis of symmetry: x = 6; vertex: (6, 36); a width of 6 meters gives the maximum possible garden area of 36 square meters. In y = −x² + 12x, a = −1 and b = 12, so the axis of symmetry is x = −12/(2 · −1) = 6. Substitute x = 6 to get y = −(6)² + 12(6) = −36 + 72 = 36, so the vertex is (6, 36). In the story, x = 6 meters is the width that produces the largest area, y = 36 square meters.
y-intercept: (0, 32); x-intercepts: (4, 0) and (−1/2, 0); (0, 32) means the stone starts 32 feet above the water, and (4, 0) means it hits the water after 4 seconds. Substitute x = 0 to find the y-intercept (0, 32). To find the x-intercepts, set y = 0 to get −16x² + 56x + 32 = 0 and factor out −8 to get −8(2x² − 7x − 4) = 0, or −8(2x + 1)(x − 4) = 0. Setting each factor to 0 gives x = −1/2 and x = 4, so the x-intercepts are (−1/2, 0) and (4, 0).
Vertex: (1, 36); domain: all real numbers; range: y ≤ 36. Find the x-coordinate of the vertex from x = −32/(2 · −16) = 1, and substitute x = 1 to get y = −16(1)² + 32(1) + 20 = 36, so the vertex is (1, 36). Over all real numbers, any x-value can be substituted into the quadratic, so the domain is all real numbers. Because a = −16 is negative, the parabola opens down from a maximum y-value of 36, so the range is y ≤ 36.
Because a = −2 is negative, the parabola opens down and has a maximum value of 7 rather than a minimum, and its range is y ≤ 7 rather than all real numbers. In y = −2x² + 12x − 11, the leading coefficient a = −2 is less than 0, which means the parabola opens downward. A downward-opening parabola reaches a peak at its vertex (3, 7), so 7 is the maximum y-value. Since no point on the graph has a y-value above 7, the range is y ≤ 7, while the domain is all real numbers.
Parabola A has the wider spread by 2 units; Parabola A's x-intercepts are −3 and 7 with a spread of 10, while Parabola B's x-intercepts are 1 and 9 with a spread of 8. For Parabola A, factor x² − 4x − 21 = 0 as (x − 7)(x + 3) = 0 to get x = 7 and x = −3, which are 7 − (−3) = 10 units apart. For Parabola B, factor x² − 10x + 9 = 0 as (x − 9)(x − 1) = 0 to get x = 9 and x = 1, which are 9 − 1 = 8 units apart. Comparing the spreads shows Parabola A is wider by 10 − 8 = 2 units.

Tier 3 Advanced

Maximum height of 121 feet at 1.75 seconds, or 7/4 seconds; hits the canyon floor at 4.5 seconds, or 9/2 seconds. The time to reach maximum height is t = −56/(2 · −16) = 7/4 = 1.75 seconds. Substituting t = 7/4 gives h = −16(49/16) + 56(7/4) + 72 = −49 + 98 + 72 = 121 feet. To find when it lands, set −16t² + 56t + 72 = 0 and factor out −8 to get −8(2t² − 7t − 9) = −8(2t − 9)(t + 1) = 0, which gives t = 9/2 = 4.5 seconds after rejecting t = −1.
Equation: A = −2x² + 80x; width: 20 meters; length: 40 meters; maximum area: 800 square meters. Since two sides have width x, the side parallel to the wall has length 80 − 2x, giving the area equation A = x(80 − 2x) = −2x² + 80x. The vertex occurs at x = −80/(2 · −2) = 20 meters. When the width is 20 meters, the length is 80 − 2(20) = 40 meters, and the maximum area is 20 · 40 = 800 square meters.
x-intercepts: (1, 0) and (9, 0); width at water level: 8 feet; maximum height: 16 feet. Read the points where the parabola crosses the horizontal axis y = 0 to find the x-intercepts (1, 0) and (9, 0). Subtract the x-coordinates to get the width of the arch at water level, 9 − 1 = 8 feet. Read the vertex at the peak of the graph, (5, 16), to find a maximum height of 16 feet above the water.
Amina multiplied −3 by −3 before squaring instead of squaring −3 first and then multiplying by −3; the correct vertex is (−3, 7) and the correct range is y ≤ 7. In −3x², the exponent applies only to x, not to the coefficient −3. Substituting x = −3 gives y = −3(−3)² − 18(−3) − 20 = −3(9) + 54 − 20 = −27 + 54 − 20 = 7. Thus, the vertex is (−3, 7), and since the parabola opens down, the range is y ≤ 7.
Flare A reaches the higher maximum height by 3 feet; Flare A's maximum height is 79 feet, and Flare B's maximum height is 76 feet. Flare A reaches its peak at t = −64/(−32) = 2 seconds, with a maximum height of h = −16(2)² + 64(2) + 15 = −64 + 128 + 15 = 79 feet. Flare B reaches its peak at t = −48/(−32) = 1.5 seconds, with a maximum height of h = −16(1.5)² + 48(1.5) + 40 = −36 + 72 + 40 = 76 feet. Flare A goes higher by 79 − 76 = 3 feet.

Tier 4 Challenge

Equation: A = −2x² + 60x; width x: 15 feet; total length L: 30 feet; maximum total area: 450 square feet. Four segments of length x and two segments of length L total 120 feet, so 4x + 2L = 120, which simplifies to L = 60 − 2x. The total area is A = x(60 − 2x) = −2x² + 60x. The vertex occurs at x = −60/(2 · −2) = 15 feet, giving a total length of L = 60 − 2(15) = 30 feet and a maximum area of 15 · 30 = 450 square feet.
Time of maximum height: 2.5 seconds, or 5/2 seconds; landing time: 5 seconds; initial upward velocity v: 80 feet per second; maximum height: 100 feet. Because (2, 96) and (3, 96) have the same height, the axis of symmetry lies halfway between t = 2 and t = 3 at t = 2.5 seconds. Since the projectile starts on the ground at (0, 0), its mirror landing point across t = 2.5 is (5, 0), so it lands at 5 seconds. Substituting (2, 96) into h = −16t² + vt gives 96 = −64 + 2v, so v = 80 feet per second, and evaluating h = −16(2.5)² + 80(2.5) gives a maximum height of 100 feet.
Dividing 12 by −4 gives a = −3, not a = 3; a parabola with a positive leading coefficient a = 3 opens up and has a minimum rather than a maximum; the correct function is y = −3x² + 18x − 15. From 12 = −4a, dividing both sides by −4 gives a = −3. Any quadratic with a positive leading coefficient opens upward and has a minimum, so a function with a maximum value of 12 must have a negative leading coefficient. Expanding y = −3(x − 1)(x − 5) = −3(x² − 6x + 5) gives y = −3x² + 18x − 15.
Parabola B has the higher maximum value, while Parabola A has the wider spread between its x-intercepts; Parabola A has maximum value 4k² and spread 4k, while Parabola B has maximum value 9k² and spread 3k. For Parabola A, the vertex is at x = −4k/(−2) = 2k with maximum y = −(2k)² + 4k(2k) = 4k², and factoring −x(x − 4k) = 0 gives x-intercepts 0 and 4k with spread 4k. For Parabola B, the vertex is at x = −12k/(−8) = 3k/2 with maximum y = −4(9k²/4) + 12k(3k/2) = 9k², and factoring −4x(x − 3k) = 0 gives x-intercepts 0 and 3k with spread 3k. Since k > 0, 9k² > 4k² and 4k > 3k.
The x-intercepts from least to greatest are x = −2, x = 0, and x = 4; the graph is above the x-axis between x = −2 and x = 0, and below the x-axis between x = 0 and x = 4. Factor out the common factor x to get y = x(x² − 2x − 8), then factor the trinomial to get y = x(x − 4)(x + 2). Setting y = 0 gives the three x-intercepts x = −2, x = 0, and x = 4. Testing x = −1 between −2 and 0 gives y = (−1)(−5)(1) = 5 > 0, so the graph is above the x-axis on (−2, 0), and testing x = 1 between 0 and 4 gives y = (1)(−3)(3) = −9 < 0, so the graph is below the x-axis on (0, 4).

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