Section 3.1

Quadratic Functions; Parabolas

Copyright ©2013 Pearson Education, Inc.

Objectives

Determine if a function is quadratic Determine if the graph of a quadratic function is a parabola that

opens down Determine if the vertex of the graph of a quadratic function is a

maximum or a minimum Determine if a function increases or decreases over a given

interval Find the vertex of the graph of a quadratic function Graph a quadratic function Write the equation of a quadratic function given information

about its graph Find the vertex form of the equation of a quadratic function

Graph of a Quadratic Function

The general form of a quadratic function is f(x) = ax2 + bx + c where a, b, and c are real numbers with a ≠ 0.

The graph of the quadratic function f(x) = ax2 + bx + c is a parabola with a “turning point” called the vertex. •The parabola opens upward (is concave up) if a is positive and the vertex is a minimum point.•The parabola opens downward (is concave down) if a is negative and the vertex is a maximum point.

Increasing and Decreasing FunctionsA function f is increasing on an interval if, for any x1 and x2 in the interval, when x2 > x1, it is true that f(x2) > f(x1).

A function f is decreasing on an interval if, for any x1 and x2 in the interval, when x2 > x1, it is true that f(x2) < f (x1).

Example

Find the vertex and graph the quadratic function f(x) = –2x2 – 4x + 6.

SolutionSince the coefficient of x2 is –2 the parabola opens downward. The x-coordinate of the vertex isAnd the y-coordinate of the vertex is

Thus the vertex is (–1, 8).

41,

2 2( 2)

b

a

f(–1) = –2(–1)2 – 4(–1) + 6 = 8

Example (cont)

Find the vertex and graph the quadratic function f(x) = –2x2 – 4x + 6.

SolutionThe x-intercepts can be found by setting f(x) = 0 and solving for x:

22 4 6 0x x

22 2 3 0x x 2 3 1 0x x

3 or 1x x

Example (cont)

SolutionThe y-intercepts can be found by computing f(0).

The axis of symmetry is the vertical line x = –1. 20 2 0 4 0 6 6f

Example—Spreadsheet solution

The graph of the function f(x) = –2x2 – 4x + 6 using Excel.

Example

A ball is thrown upward at 64 feet per second from the top of an 60-foot-high building.a. Write the quadratic function that models the height (in feet) of the ball as a function of the time t (in seconds).

Solutiona. The model has the form S(t) = –16t2 + v0t + h0, where v0 = 64 and h0 = 60. Thus the model is

S(t) = –16t2 + 64t + 60 (feet)

Example (cont)

b. Find the t-coordinate and S-coordinate of the vertex of the graph of this quadratic function.

Solutionb. The height S is the function of time t, and the

t-coordinate of the vertex is

The S coordinate of the vertex is the value of S at t = 2, so S = –16(–2)2 + 64(2) + 60 = 124.

The vertex is (2, 124).

64

2.2 2 16

bt

a

Example (cont)

c. Graph the model.

Solution

X

Y

0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

15

30

45

60

75

90

105

120

135

150

0

S = –16t2 + 64t + 60

Example (cont)

d. Explain the meaning of the coordinates of the vertex for this model.

Solutiond. The graph is a parabola that opens down, so the

vertex is the highest point on the graph and the function has a minimum there. The t-coordinate of the vertex, 2, is the time (in seconds) at which the ball reaches the maximum height, and the S-coordinate, 124, is the maximum height (in feet) that the ball reaches.

Graph of a Quadratic FunctionIn general, the graph of the function

y = a(x h)2 + k is a parabola with its vertex at the point (h, k).

The parabola opens upward if a > 0, and the vertex is a minimum.

The parabola opens downward if a < 0, and the vertex is a maximum.

The axis of symmetry of the parabola has equation x = h.The a is the same as the leading coefficient in

y = ax2 + bx + c, so the larger the value of |a|, the narrower the parabola will be.

Example

Right Sports Management had its monthly maximum profit, $450,000, when it produced and sold 5500 Waist Trimmers. Its fixed cost is $155,000. If the profit can be modeled by a quadratic function of x, the number of Waist Trimmers produced and sold each month, find this quadratic function P(x).

SolutionWhen 0 units are produced, the cost is $155,000 and the revenue is $0. Thus, the profit is –$155,000 when 0 units are produced, and the y-intercept of the graph of the function is (0, –155,000).

Example (cont)

Right Sports Management had its monthly maximum profit, $450,000, when it produced and sold 5500 Waist Trimmers. Its fixed cost is $155,000. If the profit can bemodeled by a quadratic function of x, the number of Waist Trimmers produced and sold each month, find this quadratic function P(x).

SolutionThe vertex of the graph of the quadratic function is (5500, 450,000). Using these points gives

P(x) = a(x 5500)2 + 450,000and –155,000 = a(0 – 5500)2 + 450,000which gives a = –0.02Thus, the quadratic function that models the profit is P (x) = –0.02(x – 5500)2 + 450,000, or P (x) = –0.02x2 + 220x – 155,000, where P(x) is in dollars and x is the number of units produced and sold.

Example

If the points in the table lie on a parabola, write the equations whose graph is the parabola.

SolutionThe vertex of the parabola is (3, 5).

The equation is y = (x – 3)2 – 5.

x 1 0 1 2

y 11 4 1 4

X

Y

-10 -8 -6 -4 -2 2 4 6 8 10

-10

-8

-6

-4

-2

2

4

6

8

10

0

Example

Write the vertex form of the equation of the quadratic function from the general form y = 2x2 – 8x + 3.

SolutionThe vertex is at and

y = 2(22) – 8(2) + 3 = –5.

Thus, the vertex form of the equation is y = 2(x – 2)2 – 5.

82,

2 2 2

bx

a