holt mcdougal algebra 1 8-3 graphing quadratic functions graph a quadratic function in the form y =...

Holt McDougal Algebra 1

8-3 Graphing Quadratic Functions

Graph a quadratic function in the form y = ax2 + bx + c.

Objective



PAGE 538Recall that a y-intercept is the y-coordinate of the point where a graph intersects the y-axis. The x-coordinate of this point is always 0. For a quadratic function written in the form y = ax2 + bx + c, when x = 0, y = c. So the y-intercept of a quadratic function is c.



Example 1: Graphing a Quadratic Function Graph y = 3x2 – 6x + 1.

Step 1 Find the axis of symmetry.

= 1

The axis of symmetry is x = 1. Simplify.

Use x = . Substitute 3

for a and –6 for b.

Step 2 Find the vertex.y = 3x2 – 6x + 1 = 3(1)2 – 6(1) + 1= 3 – 6 + 1= –2

The vertex is (1, –2).

The x-coordinate of the vertex is 1. Substitute 1 for x.

Simplify.The y-coordinate is –2.



Example 1 Continued

Step 3 Find the y-intercept.

y = 3x2 – 6x + 1

y = 3x2 – 6x + 1

The y-intercept is 1; the graph passes through (0, 1).

Identify c.



Step 4 Find two more points on the same side of the axis of symmetry as the point containing the y-intercept.

Since the axis of symmetry is x = 1, choose x-values less than 1.

Let x = –1.

y = 3(–1)2 – 6(–1) + 1

= 3 + 6 + 1 = 10

Let x = –2.

y = 3(–2)2 – 6(–2) + 1

= 12 + 12 + 1 = 25

Substitutex-coordinates.

Simplify.

Two other points are (–1, 10) and (–2, 25).

Example 1 Continued



Graph y = 3x2 – 6x + 1.Step 5 Graph the axis of symmetry, the vertex, the point containing the y-intercept, and two other points.

Step 6 Reflect the points across the axis of symmetry. Connect the points with a smooth curve.

Example 1 Continued

x = 1(–2, 25)

(–1, 10)

(0, 1)(1, –2)

x = 1

(–1, 10)

(0, 1)

(1, –2)

(–2, 25)



Because a parabola is symmetrical, each point is the same number of units away from the axis of symmetry as its reflected point.

Helpful Hint



Check It Out! Example 1b

Graph the quadratic function.

y + 6x = x2 + 9


Simplify.

Use x = . Substitute 1

for a and –6 for b.

The axis of symmetry is x = 3.

= 3

y = x2 – 6x + 9 Rewrite in standard form.



Step 2 Find the vertex.

Simplify.

Check It Out! Example 1b Continued

= 9 – 18 + 9

= 0

The vertex is (3, 0).

The x-coordinate of the vertex is 3. Substitute 3 for x.

The y-coordinate is 0.

y = x2 – 6x + 9

y = 32 – 6(3) + 9




y = x2 – 6x + 9

y = x2 – 6x + 9


Identify c.




Step 4 Find two more points on the same side of the axis of symmetry as the point containing the y- intercept.

Since the axis of symmetry is x = 3, choose x-values less than 3.

Let x = 2y = 1(2)2 – 6(2) + 9

= 4 – 12 + 9 = 1

Let x = 1 y = 1(1)2 – 6(1) + 9

= 1 – 6 + 9

= 4

Substitutex-coordinates.

Simplify.

Two other points are (2, 1) and (1, 4).




Step 5 Graph the axis of symmetry, the vertex, the point containing the y-intercept, and two other points.

Step 6 Reflect the points across the axis of symmetry. Connect the points with a smooth curve.

y = x2 – 6x + 9


x = 3

(3, 0)

(0, 9)

(2, 1)

(1, 4)

(0, 9)

(1, 4)

(2, 1)

x = 3

(3, 0)



Check It Out! Example 2

As Molly dives into her pool, her height in feet above the water can be modeled by the function f(x) = –16x2 + 16x + 12, where x is the time in seconds after she begins diving. Find the maximum height of her dive and the time it takes Molly to reach this height. Then find how long it takes her to reach the pool.




Use x = . Substitute

–16 for a and 16 for b.

Simplify.

The axis of symmetry is x = 0.5.

Check It Out! Example 2 Continued



Step 2 Find the vertex.

f(x) = –16x2 + 16x + 12

= –16(0.5)2 + 16(0.5) +12

= –16(0.25) + 8 +12

= –4 + 20

= 16

The vertex is (0.5, 16).

Simplify.

The y-coordinate is 9.

The x-coordinate of the vertex is 0.5. Substitute 0.5 for x.





Identify c.f(x) = –16x2 + 16x + 12





Step 4 Find another point on the same side of the axis of symmetry as the point containing the y-intercept.

Since the axis of symmetry is x = 0.5, choose an x-value that is less than 0.5.

Let x = 0.25

f(x) = –16(0.25)2 + 16(0.25) + 12

= –1 + 4 + 12

= 15 Another point is (0.25, 15).

Substitute 0.25 for x.

Simplify.




Step 5 Graph the axis of symmetry, the vertex, the point containing the y-intercept, and the other point. Then reflect the points across the axis of symmetry. Connect the points with a smooth curve.




The vertex is (0.5, 16). So at 0.5 seconds, Molly's dive has reached its maximum height of 16 feet. The graph shows the zeros of the function are –0.5 and 1.5 seconds. At –0.5 seconds the dive has not begun, and at 1.5 seconds she reaches the pool. Molly reaches the pool in 1.5 seconds.




Look Back44

Check by substituting (0.5, 16) and (1.5, 0) into the function.

16 = 16


0 = 0

16 = –16(0.5)2 + 16(0.5) + 12?

16 = –4 + 8 +12?

0 = –16(1.5)2 + 16(1.5) +12?

0 = –36 + 24 + 12?

holt mcdougal algebra 1 8-3 graphing quadratic functions graph a quadratic function in the form y =...

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