1 topic 7.4.2 drawing graphs of quadratic functions drawing graphs of quadratic functions

1

Topic 7.4.2Topic 7.4.2

Drawing Graphs ofQuadratic FunctionsDrawing Graphs of

Quadratic Functions

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Topic7.4.2

Drawing Graphs of Quadratic FunctionsDrawing Graphs of Quadratic Functions

California Standards:21.0 Students graph quadratic functions and know that their roots are the x-intercepts.

22.0 Students use the quadratic formula or factoring techniques or both to determine whether the graph of a quadratic function will intersect the x-axis in zero, one, or two points.

What it means for you:You’ll graph quadratic functions by finding their roots.

Key words:• quadratic• parabola• intercept• vertex• line of symmetry• root

3

Topic7.4.2


In this Topic you’ll use methods for finding the intercepts and the vertex of a graph to draw graphs of quadratic functions.

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Topic7.4.2

Find the Roots of the Corresponding Equations


In general, a good way to graph the function y = ax2 + bx + c is to find:

(iii) the vertex.

(ii) the y-intercept — this involves setting x = 0,

(i) the x-intercepts (if there are any) — this involves solving a quadratic equation,

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Topic7.4.2

Example 1

Solution follows…


Sketch the graphs of y = x2 – 3x + 2 and y = –2x2 + 6x – 4.

Solution

(i) To find the x-intercepts of the graph of y = x2 – 3x + 2, you need to solve: x2 – 3x + 2 = 0

So the x-intercepts are (1, 0) and (2, 0).

Using the zero property, x = 1 or x = 2.

This quadratic factors to give: (x – 1)(x – 2) = 0 –2 0 2 4

0

2

4

6y

x

–2

–4

–6

Solution continues…

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Topic7.4.2

Example 1



Solution (continued)

(ii) To find the y-intercept, put x = 0 into y = x2 – 3x + 2. This gives y = 2, so the y-intercept is at (0, 2).

–2 0 2 40

2

4

6y

x

–2

–4

–6


(iii) The x-coordinate of the vertex is always halfway between the x-intercepts.

So the x-coordinate of the vertex

is given by: x = =3

2

1 + 2

2

7

And the y-coordinate of the vertex is:

Topic7.4.2

Example 1




–2 0 2 40

2

4

6y

x

–2

–4

–6


Also, the parabola’s line of symmetry passes through the vertex.

So, the line of symmetry is the line x = .3

2

So the vertex is at , – .3

2

1

4

3

2

3

2

1

4

2– 3 × + 2 = –

y = x2 – 3x + 2

8

The next function is the same as in the previous example, only multiplied by –2.

Topic7.4.2

Example 1




–2 0 2 40

2

4

6y

x

–2

–4

–6


y = x2 – 3x + 2

The coefficient of x2 is negative this time, so the graph is concave down.

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Topic7.4.2

Example 1




–2 0 2 40

2

4

6y

x

–2

–4

–6


y = x2 – 3x + 2(i) To find the x-intercepts of the graph of y = –2x2 + 6x – 4, you need to solve: –2x2 + 6x – 4 = 0

This quadratic factors to give: –2(x – 1)(x – 2) = 0.

This means the x-intercepts are at:(1, 0) and (2, 0).

Using the zero property, x = 1 or x = 2.

10

Topic7.4.2

Example 1




–2 0 2 40

2

4

6y

x

–2

–4

–6


y = x2 – 3x + 2(ii) Put x = 0 into y = –2x2 + 6x – 4 to find the y-intercept.

The y-intercept is (0, –4).

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Topic7.4.2

Example 1




–2 0 2 40

2

4

6y

x

–2

–4

–6

y = x2 – 3x + 2(iii) The vertex is at x = . 3

2

3

2

3

2

1

2

2–2 × + 6 × – 4 =

So the y-coordinate of the vertex is at:

and, the line of symmetry is the line x = .3

2

The coordinates of the vertex are , ,3

2

1

2

y = –2x2 + 6x – 4

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1. Find the x–intercepts (if there are any).

2. Find the y–intercepts (if there are any).

3. Find the vertex.

4. Using the vertex, x-intercepts, and y-intercepts, graph the quadratic.

Topic7.4.2

Guided Practice

Solution follows…


Exercises 1–4 are about the quadratic y = x2 – 1.

Let y = 0 and factor: 0 = (x – 1)(x + 1) so x = 1 or x = –1. So, the x-intercepts are (1, 0) and (–1, 0).

When x = 0, y = 0 – 1 = – 1. So, the y-intercept is (0, –1).

x-coordinate: [1 + (–1)] ÷ 2 = 0.y-coordinate: y = 0 – 1 = –1. So, the vertex is at (0, –1). –6 –4 –2 0 2 4 60

2

4

6y

x

–2

–4

–6

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5. Find the x–intercepts (if there are any).

6. Find the y–intercepts (if there are any).

7. Find the vertex.

8. Using the vertex, x-intercepts, and y-intercepts, graph the quadratic.

Topic7.4.2

Guided Practice

Solution follows…


Exercises 5–8 are about the quadratic y = (x – 1)2 – 4.

Rearrange to form a standard quadratic: y = x2 – 2x – 3 Let y = 0 and factor: 0 = (x – 3)(x + 1) so x = 3 or x = –1. So, the x-intercepts are (3, 0) and (–1, 0).

When x = 0, y = (0 – 1)2 – 4 = 1 – 4 = –3. So, the y-intercept is (0, –3).

x-coordinate: [3 + (–1)] ÷ 2 = 1.y-coordinate: y = (1 – 1)2 – 4 = –4.

So, the vertex is at (1, –4).

–6 –4 –2 0 2 4 60

2

4

6y

x

–2

–4

–6

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Topic7.4.2

Independent Practice

Solution follows…


For each of the quadratics in Exercises 1–2, follow these steps: i) Find the x–intercepts (if any), ii) Find the y–intercepts (if any), iii) Find the vertex, iv) Using the vertex, x-intercepts,

and y-intercepts, graph the quadratic.

1. y = x2 – 2x

2. y = x2 + 2x – 3

x-intercepts: (0, 0) and (2, 0)y-intercept: (0, 0)vertex: (1, –1)

x-intercepts: (–3, 0) and (1, 0)y-intercept: (0, –3)vertex: (–1, –4)

–6 –4 –2 0 2 4 60

2

4

6y

x

–2

–4

–6

12

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Topic7.4.2


Solution follows…




3. y = –4x2 – 4x + 3

4. y = x2 – 4x-intercepts: (–2, 0) and (2, 0)y-intercept: (0, –4)vertex: (0, –4)

–6 –4 –2 0 2 4 60

2

4

6y

x

–2

–4

–6

34

x-intercepts: (– , 0) and ( , 0)

y-intercept: (0, 3), vertex: (– , 4)

3

2

1

21

2

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Topic7.4.2


Solution follows…




5. y = x2 + 4x + 4

6. y = –x2 + 4x + 5x-intercepts: (5, 0) and (–1, 0)y-intercept: (0, 5)vertex: (2, 9)

56

x-intercept: (–2, 0)y-intercept: (0, 4)vertex: (–2, 0)

–6 –4 –2 0 2 4 6

4

6

8

10y

x

2

0

–2

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Topic7.4.2


Solution follows…


For the quadratics in Exercises 7, follow these steps: i) Find the x–intercepts (if any), ii) Find the y–intercepts (if any), iii) Find the vertex, iv) Using the vertex, x-intercepts,


7. y = –9x2 – 6x + 3

–6 –4 –2 0 2 4 60

2

4

6y

x

–2

–4

–6

7

x-intercepts: (–1 , 0) and ( , 0)

y-intercept: (0, 3), vertex: (– , 4)

1

31

3

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Topic7.4.2


Solution follows…


Describe the characteristics of quadratic graphs of the form y = ax2 + bx + c that have the following features, or say if they are not possible.

8. No x-intercepts

9. One x-intercept

10. Two x-intercepts

11. Three x-intercepts

The graph is either concave up with the vertex above the x–axis, or concave down with the vertex below the x–axis.

Not possible.

The graph is either concave up with the vertex below the x–axis, or concave down with the vertex above the x–axis.

The vertex is the x–intercept.

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Topic7.4.2


Solution follows…


Describe the characteristics of quadratic graphs of the form y = ax2 + bx + c that have the following features, or say if they are not possible.

12. No y-intercepts

13. One y-intercept

14. Two y-intercepts

15. Three y-intercepts

All quadratic equations of the form y = ax2 + bx + c will have one y–intercept.

Not possible.

Not possible.

Not possible.

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Topic7.4.2


Solution follows…


17. Which quadratic equation has the following features? Vertex (0, 16), x-intercepts (4, 0), (–4, 0), and y-intercept (0, 16)

16. Which quadratic equation has the following features? Vertex (3, –4), x-intercepts (1, 0), (5, 0), and y-intercept (0, 5)

y = (x – 3)2 – 4 or y = x2 – 6x + 5

y = –x2 + 16

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Topic7.4.2

Round UpRound Up


A quadratic function has the general form y = ax2 + bx + c (where a 0).

When you draw the graph of a quadratic, the value of a determines whether the parabola is concave up (u-shaped) or concave down (n-shaped), and how steep it is.

Changing the value of c moves the graph in the direction of the y-axis.

Note that if a = 0, the function becomes y = bx + c, which is a linear function whose graph is a straight line.

1 topic 7.4.2 drawing graphs of quadratic functions drawing graphs of quadratic functions

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