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Warm Up Section 3.3 (1). Solve: 2 x – 3 = 12 (2). Solve and graph: 3 x + 1 ≤ 7PowerPoint Presentation

Warm Up Section 3.3 (1). Solve: 2 x – 3 = 12 (2). Solve and graph: 3 x + 1 ≤ 7

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### Graphing Absolute Value Functions

(1). Solve: 2x – 3 = 12

(2). Solve and graph: 3x + 1 ≤ 7

(3). Solve and graph: 2 – x > 9

(4). {(0, 3), (1, -4), (5, 6), (4, 6)}

State domain, range.

Is relation a function?

If function, is it one-to-one?

Solutions for Warm Up Section 3.3

(1). Solve: 2x – 3 = 12

(2). Solve and graph: 3x + 1 ≤ 7

(3). Solve and graph: 2 – x > 9

(4). {(0, 3), (1, -4), (5, 6), (4, 6)}

State domain, range.

Is relation a function?

If function, is it one-to-one?

15/2 or -9/2

-8/3 2

-7 11

D = 0, 1, 5, 4

R = 3, -4, 6

Function

Not one-to-one

D: {0, 1, 2, 3, 4}, R: { 3, 1, 2, 4}, F: Yes; 1-to-1: No

D: {-2, -1, 0, 1}, R: {-3, -1, 1, 3, 5}, F: No; 1-to-1: No

D: {1, 2, 3}, R: {1, 2, 3}, F: No; 1-to-1: No

D: {-2, -1, 4, 0, 2}, R: {1}, F: Yes, 1-to-1: No

D: {1, 0, 5, 2}, R: {4, 2, -3, 0}, F: No; 1-to-1: No

D: {x: x ≤ 7 }, R: {y: y ≤ 4}, F: Yes, 1-1:No

F: Yes, 1-to-1: No 8. F: Yes, 1-to-1: No

F: Yes, 1-to-1: Yes 13. No; f(-2) = 2, f(1) = 2

14. Yes; f(2) = -3, f(-3) = 12 15. No; f(-4) = 2, f(0) = 2

Section 3.3

Standard: MM2A1 b

Essential Question: How do I graph and describe an absolute value function?

Absolute value function: f(x) = |x|

Vertex of an absolute value graph: the highest or lowest point on the graph of an absolute value function

Transformation: changes a graph’s size, shape, position or orientation

Translation: a transformation that shifts a graph horizontally and/or vertically, but doesn’t not change its size, shape or orientation

Reflection: when a = –1, the graph y = a|x| flips over the x-axis

Axis of symmetry: a vertical line that divides the graph into mirror images

Zeros: values of x that make the value of f(x) = 0

and graph of y = x

Identify each of the following:

(a). Domain: ______________

(b). Range: _______________

(c). Vertex: _______________

(d). Axis of symmetry: ______

(e). Opens: _______________

(f). Max or min: ___________

(g). x-intercept: ____________

(h). y-intercept: ___________

(i). Extrema: ______________

(j). Increasing: ____________

(k). Decreasing: ____________

(l). Rate of change: ________

All reals

y≥ 0

(0, 0)

x = 0

Upward

Min

(0, 0)

(0, 0)

Min value = 0

x≥ 0

x≤ 0

Left: -1; Right: 1

Graph y = |x + 3| + 2.

Recall:

By adding 3 to x, we are

shifting the parent graph

3 units to the left.

By adding 2 to y, we are

then shifting the graph

2 units upward.

Hence, the vertex is now at (-3, 2) and the basic shape is the

same.

1. Graph y = |x + 3| + 2.

Identify each of the following:

(a). Domain: ______________

(b). Range: _______________

(c). Vertex: _______________

(d). Axis of symmetry: ______

(e). Opens: _______________

(f). Max or min: ___________

(g). x-intercept: ____________

(h). y-intercept: ___________

(i). Extrema: ______________

(j). Increasing: ____________

(k). Decreasing: ____________

(l). Rate of change: ________

All reals

y≥ 2

(-3, 2)

x = -3

Upward

Min

None

(0, 5)

Min value = 2

x≥ -3

x≤ -3

Left: -1; Right: 1

2. Graph y = -| x– 4 | + 3.

Identify each of the following:

(a). Domain: ______________

(b). Range: _______________

(c). Vertex: _______________

(d). Axis of symmetry: ______

(e). Opens: _______________

(f). Max or min: ___________

(g). x-intercept: ____________

(h). y-intercept: ___________

(i). Extrema: ______________

(j). Increasing: ____________

(k). Decreasing: __________

(l). Rate of change: ________

All reals

y≤ 3

(4, 3)

x = 4

Downward

Max

(1, 0) and (7, 0)

(0, -1)

Max value = 3

x≤ 4

x≥ 4

Left: 1; Right: -1

y = -| x – 4 | + 3, solve

each equation or inequality:

(a). -| x – 4 | + 3 = 0

(b). -| x – 4 | + 3 > 0

(c). -| x – 4 | + 3 < 0

x = 1, 7

1 < x < 7

x < 1or x > 7

4. Graph

Identify each of the following:

(a). Domain: ______________

(b). Range: _______________

(c). Vertex: _______________

(d). Axis of symmetry: ______

(e). Opens: _______________

(f). Max or min: ___________

(g). x-intercept: ____________

(h). y-intercept: ___________

(i). Extrema: ______________

(j). Increasing: ____________

(k). Decreasing: ____________

(l). Rate of change: ________

(-∞,∞)

[-2, ∞) or y ≥ -2

(1, -2)

x = 1

Upward

Min

(5, 0) and (-3, 0)

(0, -1.5)

Min value = -2

[1, ∞) or x ≥ 1

(-∞, 1] or x ≤ 1

Left: -1/2; Right: 1/2

, solve

each equation or inequality:

(a).

(b).

(c).

x = -3, 5

x < -3or x > 5

-3 < x < 5

6. Graph y = -3|x + 2| – 5

Identify each of the following:

(a). Domain: ______________

(b). Range: _______________

(c). Vertex: _______________

(d). Axis of symmetry: ______

(e). Opens: _______________

(f). Max or min: ___________

(g). x-intercept: ____________

(h). y-intercept: ___________

(i). Extrema: ______________

(j). Increasing: ____________

(k). Decreasing: ____________

(l). Rate of change: ________

All reals

y≤ -5

(-2, -5)

x = -2

Downward

Max

None

(0, -11)

Max value = -5

x≤ -2

x≥ -2

Left: 3; Right: -3

7. Write a function for the

graph shown.

The vertex of the graph is _______________.

So the graph has the form y = a|x - ____| + _____.

To determine the value of a, compute the absolute value of the slope from (1, -2) to

(3, 2).

(3, 2)

(3, 2)

2

3

(1, -2)

m = 4/-2

2 = 2

y = -2x – 3+2

An equation for the graph is _____________________

8. Write a function for the graph shown.

Vertex: _______________.

y = a|x - ____| + _____

(4, 5)

(-2, 3)

-2

3

(-2, 3)

To determine the value of a, compute the absolute value of the slope from (-2, 3) to

(4, 5).

m = 2/6

1/3 = 1/3

An equation for the graph is _____________________

9. Without graphing, determine the

intercepts for the graph of y = 5x + 12 – 45

x-intercept: Let y = 0

0 = 5x + 12 – 45

45 = 5x + 12

9 = x + 12

x + 12 = 9

x = -3

OR

x + 12 = -9

x = -21

x-intercepts: (-3, 0) and (-21, 0)

9. Without graphing, determine the

intercepts for the graph of y = 5x + 12 – 45

y-intercept: Let x = 0

y = 50 + 12 – 45

y = 512 – 45

y = 5(12) – 45

y = 60 – 45

y = 15

y-intercept: (0, 15)

10. Without graphing, determine the values of x

for which y = 5x + 12 – 45 lies above the

x-axis.

5x + 12 – 45 > 0

5x + 12 > 45

x + 12 > 9

x + 12 > 9

x > -3

OR

x + 12 < -9

x < -21

x < -21 or x > -3

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