# GRAPHING ABSOLUTE VALUE FUNCTION AND FIND DOMAIN AND RANGE

By comparing the absolute value function with the following general form and finding out the details given below, we can draw the graph simply.

y = a|x - h| + k

Here (h, k) is the vertex or minimum or maximum point of the absolute value function

a - slope of the function

• If a is positive, the graph of the absolute value function will be in the shape of V.
• If a is negative, the graph of the absolute value function will be in the shape ∧.

Applying some inputs, we can find the outputs.

Graph the following absolute value function. State the domain and range.

Problem 1 :

f(x) = |x| - 7

Solution :

Comparing with y = a|x - h| + k

The minimum point or vertex is at (0, -7). Since a is positive the absolute value function opens up.

Slope = 1

 Inputs-2-1012 Outputs-5-6-7-6-5

(-2, -5)(-1, -6) (0, -7) (1, -6) and (2, -5)

• Domain is all real values, that is  (-∞, ∞)
• Range is [-7, ∞)

Problem 2 :

f(x) = |x - 3| + 4

Solution :

Comparing with y = a|x - h| + k

The minimum point or vertex is at (3, 4). Since a is positive the absolute value function opens up.

Slope = 1

 Inputs12345 Outputs65456

Some of the points on the absolute value function are (1, 6)(2, 5) (3, 4) (4, 5) and (5, 6)

• Domain is all real values, that is  (-∞, ∞)
• Range is [4, ∞)

Problem 3 :

f(x) = 3|x - 6|

Solution :

Comparing with y = a|x - h| + k

The minimum point or vertex is at (6, 0). Since a is positive the absolute value function opens up.

Slope = 3

 Inputs45678 Outputs63036

Some of the points are (4, 6) (5, 3) (6, 0) (7, 3) and (8, 6).

• Domain is all real values, that is  (-∞, ∞)
• Range is [0, ∞)

Problem 4 :

f(x) = -|x + 2| + 8

Solution :

Comparing with y = a|x - h| + k

The minimum point or vertex is at (-2, 8). Since a is positive the absolute value function opens down.

Slope = -1

 Inputs-3-2-101 Outputs78785

So, the points are (-3, 7) (-2, 8) (-1, 7) (0, 8) and (1, 5).

• Domain is all real values, that is  (-∞, ∞)
• Range is (-∞, 8]

Problem 5 :

f(x) = 2|-2(x - 1)| - 4

Solution :

Comparing with y = a|x - h| + k

The minimum point or vertex is at (1, -4). Since a is positive the absolute value function opens up.

Slope = 2

 Inputs-2-1012 Outputs840-40

Some of the points on the absolute value function are (-2, 8)(-1, 4) (0, 0) (1, -4) and (2, 0).

• Domain is all real values, that is  (-∞, ∞)
• Range is [-4, ∞)

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