# Graph y=2/3*|x-1|-4

y=23⋅|x-1|-4
Find the absolute value vertex. In this case, the vertex for y=2(|x-1|-6)3 is (0,2(|x-1|-6)3).
To find the x coordinate of the vertex, set the inside of the absolute value equal to 0. In this case, =0.
=0
Since there is no variable to replace, the result is 2(|x-1|-6)3.
y=2(|x-1|-6)3
The absolute value vertex is (0,2(|x-1|-6)3).
(0,2(|x-1|-6)3)
(0,2(|x-1|-6)3)
The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.
Interval Notation:
(-∞,∞)
Set-Builder Notation:
{x|x∈ℝ}
For each x value, there is one y value. Select few x values from the domain. It would be more useful to select the values so that they are around the x value of the absolute value vertex.
Substitute the x value -2 into f(x)=2(|x-1|-6)3. In this case, the point is (-2,-2).
Replace the variable x with -2 in the expression.
f(-2)=2(|(-2)-1|-6)3
Simplify the result.
Simplify the numerator.
Subtract 1 from -2.
f(-2)=2(|-3|-6)3
The absolute value is the distance between a number and zero. The distance between -3 and 0 is 3.
f(-2)=2(3-6)3
Subtract 6 from 3.
f(-2)=2⋅-33
f(-2)=2⋅-33
Simplify the expression.
Multiply 2 by -3.
f(-2)=-63
Divide -6 by 3.
f(-2)=-2
f(-2)=-2
y=-2
y=-2
y=-2
Substitute the x value -1 into f(x)=2(|x-1|-6)3. In this case, the point is (-1,-83).
Replace the variable x with -1 in the expression.
f(-1)=2(|(-1)-1|-6)3
Simplify the result.
Simplify the numerator.
Subtract 1 from -1.
f(-1)=2(|-2|-6)3
The absolute value is the distance between a number and zero. The distance between -2 and 0 is 2.
f(-1)=2(2-6)3
Subtract 6 from 2.
f(-1)=2⋅-43
f(-1)=2⋅-43
Simplify the expression.
Multiply 2 by -4.
f(-1)=-83
Move the negative in front of the fraction.
f(-1)=-83
f(-1)=-83
y=-83
y=-83
y=-83
Substitute the x value 1 into f(x)=2(|x-1|-6)3. In this case, the point is (1,-4).
Replace the variable x with 1 in the expression.
f(1)=2(|(1)-1|-6)3
Simplify the result.
Simplify the numerator.
Subtract 1 from 1.
f(1)=2(|0|-6)3
The absolute value is the distance between a number and zero. The distance between 0 and 0 is 0.
f(1)=2(0-6)3
Subtract 6 from 0.
f(1)=2⋅-63
f(1)=2⋅-63
Simplify the expression.
Multiply 2 by -6.
f(1)=-123
Divide -12 by 3.
f(1)=-4
f(1)=-4
y=-4
y=-4
y=-4
Substitute the x value 2 into f(x)=2(|x-1|-6)3. In this case, the point is (2,-103).
Replace the variable x with 2 in the expression.
f(2)=2(|(2)-1|-6)3
Simplify the result.
Simplify the numerator.
Subtract 1 from 2.
f(2)=2(|1|-6)3
The absolute value is the distance between a number and zero. The distance between 0 and 1 is 1.
f(2)=2(1-6)3
Subtract 6 from 1.
f(2)=2⋅-53
f(2)=2⋅-53
Simplify the expression.
Multiply 2 by -5.
f(2)=-103
Move the negative in front of the fraction.
f(2)=-103
f(2)=-103