Algebra unit 10.3

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UNIT 10.3 SOLVING QUADRATIC UNIT 10.3 SOLVING QUADRATIC EQUATIONS EQUATIONS

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Unit 10.3

Transcript of Algebra unit 10.3

Page 1: Algebra unit 10.3

UNIT 10.3 SOLVING QUADRATICUNIT 10.3 SOLVING QUADRATICEQUATIONSEQUATIONS

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Warm Up

Find the x-intercept of each linear function.

1. y = 2x – 3 2.

3. y = 3x + 6 Evaluate each quadratic function for the given input values.

4. y = –3x2 + x – 2, when x = 2

5. y = x2 + 2x + 3, when x = –1

–2

–12

2

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Find the zeros of a quadratic function from its graph.

Find the axis of symmetry and the vertex of a parabola.

Objectives

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zero of a functionaxis of symmetry

Vocabulary

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Recall that an x-intercept of a function is a value of x when y = 0. A zero of a function is an x-value that makes the function equal to 0. So a zero of a function is the same as an x-intercept of a function. Since a graph intersects the x-axis at the point or points containing an x-intercept, these intersections are also at the zeros of the function. A quadratic function may have one, two, or no zeros.

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Example 1A: Finding Zeros of Quadratic Functions From Graphs

Find the zeros of the quadratic function from its graph. Check your answer.

y = x2 – 2x – 3

The zeros appear to be –1 and 3.

y = (–1)2 – 2(–1) – 3 = 1 + 2 – 3 = 0 y = 32 –2(3) – 3 = 9 – 6 – 3 = 0

y = x2 – 2x – 3

Check

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Example 1B: Finding Zeros of Quadratic Functions From Graphs

Find the zeros of the quadratic function from its graph. Check your answer.

y = x2 + 8x + 16

y = (–4)2 + 8(–4) + 16 = 16 – 32 + 16 = 0

y = x2 + 8x + 16

Check

The zero appears to be –4.

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Notice that if a parabola has only one zero, the zero is the x-coordinate of the vertex.

Helpful Hint

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Example 1C: Finding Zeros of Quadratic Functions From Graphs

Find the zeros of the quadratic function from its graph. Check your answer.

y = –2x2 – 2

The graph does not cross the x-axis, so there are no zeros of this function.

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Check It Out! Example 1a

Find the zeros of the quadratic function from its graph. Check your answer.

y = –4x2 – 2

The graph does not cross the x-axis, so there are no zeros of this function.

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Check It Out! Example 1b

Find the zeros of the quadratic function from its graph. Check your answer.

y = x2 – 6x + 9

The zero appears to be 3.

y = (3)2 – 6(3) + 9 = 9 – 18 + 9 = 0

y = x2 – 6x + 9

Check

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A vertical line that divides a parabola into two symmetrical halves is the axis of symmetry. The axis of symmetry always passes through the vertex of the parabola. You can use the zeros to find the axis of symmetry.

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Example 2: Finding the Axis of Symmetry by Using Zeros

Find the axis of symmetry of each parabola.A. (–1, 0) Identify the x-coordinate

of the vertex.The axis of symmetry is x = –1.

Find the average of the zeros.

The axis of symmetry is x = 2.5.

B.

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Check It Out! Example 2

Find the axis of symmetry of each parabola.

(–3, 0) Identify the x-coordinate of the vertex.

The axis of symmetry is x = –3.

a.

b. Find the average of the zeros.

The axis of symmetry is x = 1.

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If a function has no zeros or they are difficult to identify from a graph, you can use a formula to find the axis of symmetry. The formula works for all quadratic functions.

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Example 3: Finding the Axis of Symmetry by Using the Formula

Find the axis of symmetry of the graph of y = –3x2 + 10x + 9.

Step 1. Find the values of a and b.

y = –3x2 + 10x + 9

a = –3, b = 10

Step 2. Use the formula.

The axis of symmetry is

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Check It Out! Example 3

Find the axis of symmetry of the graph of y = 2x2 + x + 3.

Step 1. Find the values of a and b.

y = 2x2 + 1x + 3a = 2, b = 1

Step 2. Use the formula.

The axis of symmetry is .

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Once you have found the axis of symmetry, you can use it to identify the vertex.

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Example 4A: Finding the Vertex of a ParabolaFind the vertex.

y = 0.25x2 + 2x + 3

Step 1 Find the x-coordinate of the vertex. The zeros are –6 and –2.

Step 2 Find the corresponding y-coordinate.y = 0.25x2 + 2x + 3

= 0.25(–4)2 + 2(–4) + 3 = –1 Step 3 Write the ordered pair.

(–4, –1)

Use the function rule.

Substitute –4 for x .

The vertex is (–4, –1).

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Example 4B: Finding the Vertex of a ParabolaFind the vertex.

y = –3x2 + 6x – 7

Step 1 Find the x-coordinate of the vertex.

a = –3, b = 10 Identify a and b.

Substitute –3 for a and 6 for b.

The x-coordinate of the vertex is 1.

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Example 4B Continued

Find the vertex.

Step 2 Find the corresponding y-coordinate.

y = –3x2 + 6x – 7 = –3(1)2 + 6(1) – 7

= –3 + 6 – 7 = –4

Use the function rule.

Substitute 1 for x.

Step 3 Write the ordered pair.

The vertex is (1, –4).

y = –3x2 + 6x – 7

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Find the vertex.

y = x2 – 4x – 10

Step 1 Find the x-coordinate of the vertex.

a = 1, b = –4 Identify a and b.

Substitute 1 for a and –4 for b.

The x-coordinate of the vertex is 2.

Check It Out! Example 4

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Find the vertex.

Step 2 Find the corresponding y-coordinate.

y = x2 – 4x – 10= (2)2 – 4(2) – 10

= 4 – 8 – 10 = –14

Use the function rule.

Substitute 2 for x.

Step 3 Write the ordered pair.

The vertex is (2, –14).

y = x2 – 4x – 10

Check It Out! Example 4 Continued

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Example 5: Application

The graph of f(x) = –0.06x2 + 0.6x + 10.26 can be used to model the height in meters of an arch support for a bridge, where the x-axis represents the water level and x represents the distance in meters from where the arch support enters the water. Can a sailboat that is 14 meters tall pass under the bridge? Explain.

The vertex represents the highest point of the arch support.

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Example 5 ContinuedStep 1 Find the x-coordinate.

a = – 0.06, b = 0.6 Identify a and b.

Substitute –0.06 for a and 0.6 for b.

Step 2 Find the corresponding y-coordinate.

= –0.06(5)2 + 0.6(5) + 10.26 f(x) = –0.06x2 + 0.6x + 10.26

= 11.76

Use the function rule.

Substitute 5 for x.

Since the height of each support is 11.76 m, the sailboat cannot pass under the bridge.

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Check It Out! Example 5 The height of a small rise in a roller coaster track is modeled by f(x) = –0.07x2 + 0.42x + 6.37, where x is the distance in feet from a supported pole at ground level. Find the height of the rise.

Step 1 Find the x-coordinate.

a = – 0.07, b= 0.42 Identify a and b.

Substitute –0.07 for a and 0.42 for b.

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Check It Out! Example 5 Continued

Step 2 Find the corresponding y-coordinate.

= –0.07(3)2 + 0.42(3) + 6.37

f(x) = –0.07x2 + 0.42x + 6.37

= 7 ft

Use the function rule.

Substitute 3 for x.

The height of the rise is 7 ft.

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Lesson Quiz: Part I

1. Find the zeros and the axis of symmetry of the parabola.

2. Find the axis of symmetry and the vertex of the graph of y = 3x2 + 12x + 8.

zeros: –6, 2; x = –2

x = –2; (–2, –4)

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Lesson Quiz: Part II

25 feet

3. The graph of f(x) = –0.01x2 + x can be used to model the height in feet of a curved arch support for a bridge, where the x-axis represents the water level and x represents the distance in feet from where the arch support enters the water. Find the height of the highest point of the bridge.

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