y=23⋅|x-1|-4

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}

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

The final answer is -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

The final answer is -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

The final answer is -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

The final answer is -103.

y=-103

y=-103

y=-103

The absolute value can be graphed using the points around the vertex (0,2(|x-1|-6)3),(-2,-2),(-1,-2.67),(1,-4),(2,-3.33)

xy-2-2-1-2.6702(|x-1|-6)31-42-3.33

xy-2-2-1-2.6702(|x-1|-6)31-42-3.33

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