# ABSOLUTE VLAUE FUNCTION REVIEW WORKSHEET

Problem 1 :

What is the vertex: y = 3|x - 1| + 2

a) (1, -2)    b) (1, 2)

Solution :

y = 3|x - 1| + 2

By comparing the given absolute value function with

y = a|x - h| + k

Vertex (h, k):

(1, 2)

Problem 2 :

What is the vertex: y = 2|x|

a) (2, 0)    b) (0, 0)

Solution :

y = 2|x|

By comparing the given absolute value function with

y = a|x - h| + k

Vertex (h, k):

(0, 0)

Problem 3 :

What is the vertex: y = |x| + 5

a) (5, 0)    b) (0, 5)

Solution :

y = |x| + 5

By comparing the given absolute value function with

y = a|x - h| + k

Vertex (h, k) :

(0, 5)

Problem 4 :

The graph of y = -2|x + 1| - 3 is

a) Vertically stretched   b) Vertical shrink

Solution :

Comparing the given equation with y = a |x - h| + k

a = -2, since the sign of a is negative, it is reflection and the value of a is 2.

a > 1, it is vertically shrink.

Problem 5 :

The graph of y = -3/5|x + 3| + 10 is

a) Vertically stretched   b) Vertical shrink

Solution :

Comparing the given equation with y = a |x - h| + k

a = -3/5, since the sign of a is negative, it is reflection and the value of a is 3/5.

0 < a < 1, it is vertically stretch.

Problem 6 :

The graph of y = 15|x| is

a) Vertically stretched   b) Vertical shrink

Solution :

Here a = 15

a > 1, it is vertically shrink.

Problem 7 :

The graph of y = 5/3|x + 2| - 1 is

a) Vertically stretched   b) Vertical shrink

Solution :

a = 5/3

a > 1, it is vertically shrink.

Problem 8 :

The graph of y = 5/3|x + 2| - 1

a) Opens up   b) Opens down

Solution :

y = a|x - h| + k

y = 5/3|x + 2| - 1

a = 5/3

It is positive. So opens up.

Problem 9 :

The graph of y = -2|x + 1| - 3

a) Opens up   b) Opens down

Solution:

y = a|x - h| + k

y = -2|x + 1| - 3

a = -2

It is negative. So opens down.

Problem 10 :

The graph of: y = 3|x| - 2 is

a) Translated vertically   b) Translated horizontally

Solution :

Comparing the given function y = 3|x| - 2 with the parent function y = |x|, for k we have -2. So, the graph is moved towards down 2 units. So translated vertically is the answer.

Problem 11 :

The graph of: y = 3|x| - 2 is

a) Translated left 2   b) Translated down 2

Solution :

Since the value of k is -2, we have to do translation 2 units down.

Problem 12 :

In order for the graph to be a vertical shrink, what will be the value of a

a)  -1 < a < 1       b) a > 1

Solution :

If a > 1, then it is vertical shrink.

Problem 13:

Domain of: y = -2|x + 1| - 3

a)  (-∞, ∞)       b) -∞ < x < 1

Solution :

There is no restriction for x values for the function given above, so

Domain: (-∞, ∞) x € R

All real numbers are domain.

Problem 14 :

Range of: y = 3|x| - 1

a)  (-∞, 3]     b) [-1, ∞),

Solution :

y = 3|x| - 1

Range: [-1, ∞), y ≥ -1

Problem 15 :

Range of: y = -|x| - 2

a)  [-2,-∞)     b) [-2, ∞)

Solution :

y = -|x| - 2

Range: [-2,-∞), y ≤ -2

Problem 16 :

Range of: y = -|x - 5| + 4

a)  (-∞, 4]     b) [4, ∞)

Solution :

y = -|x - 5| + 4

Range: [4, ∞), y ≥ 4

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