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
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 -1 0 1 2 |
Outputs -5 -6 -7 -6 -5 |
(-2, -5)(-1, -6) (0, -7) (1, -6) and (2, -5)
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
Inputs 1 2 3 4 5 |
Outputs 6 5 4 5 6 |
Some of the points on the absolute value function are (1, 6)(2, 5) (3, 4) (4, 5) and (5, 6)
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
Inputs 4 5 6 7 8 |
Outputs 6 3 0 3 6 |
Some of the points are (4, 6) (5, 3) (6, 0) (7, 3) and (8, 6).
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 -1 0 1 |
Outputs 7 8 7 8 5 |
So, the points are (-3, 7) (-2, 8) (-1, 7) (0, 8) and (1, 5).
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 -1 0 1 2 |
Outputs 8 4 0 -4 0 |
Some of the points on the absolute value function are (-2, 8)(-1, 4) (0, 0) (1, -4) and (2, 0).
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