To find increasing and decreasing interval of absolute value function, first we should find the vertex. Because vertex is the turning point of the curve.
We know that graphing form of absolute value function will have the shape V. It may be V or inverted V, based on the sign of coefficient of absolute value function, we will decide.
If the curve opens up,
If the curve opens down,
To get more examples to know how to graph absolute value function, please visit the pages given.
Problem 1 :
f(x) = -3│x - 4│ + 3
Solution :
a = -3, h = 4, k = 3
Vertex:
f(x) = -3│x - 4│ + 3
Comparing with f(x) = a |x - h|+ k
Vertex (h, k) = (4, 3)
Slope :
Slope (a) = -3
The curve will open down.
y- intercept :
y- Intercept, put x = 0
y = -3│0 - 4│ + 3
y = -3(4) + 3
y = -12 + 3
y = -9
y- Intercept = (0, -9)
Zeros :
x- Intercept, put y = 0
-3│x - 4│ + 3 = 0
-3│x - 4│= -3
│x - 4│= -3/-3
│x - 4│= 1
x = 1 + 4
x = 5
x- Intercept is (5, 0)
Domain and range :
Increasing and decreasing :
Problem 2 :
f(x) = -1/2│x - 2│ + 4
Solution :
a = -1/2, h = 2, k = 4
Vertex :
f(x) = -1/2│x - 2│ + 4
Comparing with f(x) = a│x - h│+ k
Vertex (h, k) = (2, 4)
Slope :
Slope (a) = -1/2
The curve will open down.
y- intercept :
y- Intercept, put x = 0
y = -1/2│0 - 2│ + 4
y = -1/2(2) + 4
y = -1 + 4
y = 3
y- Intercept = (0, 3)
Zeros :
x- Intercept, put y = 0
-1/2│x - 2│ + 4 = 0
-1/2│x - 2│= -4
│x - 2│= -4(-2)
│x - 2│= 8
x = 8 + 2
x = 10
x- Intercept is (10, 0)
Domain and range:
Increasing and decreasing:
Problem 3 :
f(x) = │x - 3│ - 2
Solution :
a = 1, h = 3, k = -2
Vertex :
f(x) = │x - 3│ - 2
Comparing with f(x) = a│x - h│+ k
Vertex (h, k) is (3, -2)
Slope :
Slope (a) = 1
The curve will open up.
y- intercept :
y- Intercept, put x = 0
y = │0 - 3│ - 2
y = 3 - 2
y = 1
y- Intercept = (0, 1)
Zeros :
x- Intercept, put y = 0
│x - 3│ - 2 = 0
│x - 3│= 2
x = 2 + 3
x = 5
x- Intercept = (5, 0)
Domain and range :
Increasing and decreasing :
Dec 08, 23 08:03 AM
Dec 08, 23 07:32 AM
Dec 08, 23 07:10 AM