Algebra Examples

y = | x - 1 | + 4

$y = | x - 1 | + 4$

Step 1

Find the absolute value vertex. In this case, the vertex for

y = | x - 1 | + 4

$y = | x - 1 | + 4$ is

(1, 4)

$(1, 4)$ .

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Step 1.1

To find the

x

$x$ coordinate of the vertex, set the inside of the absolute value

x - 1

$x - 1$ equal to

0

$0$ . In this case,

x - 1 = 0

$x - 1 = 0$ .

x - 1 = 0

$x - 1 = 0$

Step 1.2

Add

1

$1$ to both sides of the equation.

x = 1

$x = 1$

Step 1.3

Replace the variable

x

$x$ with

1

$1$ in the expression.

y = | (1) - 1 | + 4

$y = | (1) - 1 | + 4$

Step 1.4

Simplify

| (1) - 1 | + 4

$| (1) - 1 | + 4$ .

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Step 1.4.1

Simplify each term.

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Step 1.4.1.1

Subtract

1

$1$ from

1

$1$ .

y = | 0 | + 4

$y = | 0 | + 4$

Step 1.4.1.2

The absolute value is the distance between a number and zero. The distance between

0

$0$ and

0

$0$ is

0

$0$ .

y = 0 + 4

$y = 0 + 4$

y = 0 + 4

$y = 0 + 4$

Step 1.4.2

Add

0

$0$ and

4

$4$ .

y = 4

$y = 4$

y = 4

$y = 4$

Step 1.5

The absolute value vertex is

(1, 4)

$(1, 4)$ .

(1, 4)

$(1, 4)$

(1, 4)

$(1, 4)$

Step 2

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:

(- \infty, \infty)

$(- \infty, \infty)$

Set-Builder Notation:

{x | x \in ℝ}

${x | x \in ℝ}$

Step 3

For each

x

$x$ value, there is one

y

$y$ value. Select a few

x

$x$ values from the domain. It would be more useful to select the values so that they are around the

x

$x$ value of the absolute value vertex.

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Step 3.1

Substitute the

x

$x$ value

- 1

$- 1$ into

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

$f (x) = | x - 1 | + 4$ . In this case, the point is

(- 1, 6)

$(- 1, 6)$ .

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Step 3.1.1

Replace the variable

x

$x$ with

- 1

$- 1$ in the expression.

f (- 1) = | (- 1) - 1 | + 4

$f (- 1) = | (- 1) - 1 | + 4$

Step 3.1.2

Simplify the result.

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Step 3.1.2.1

Simplify each term.

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Step 3.1.2.1.1

Subtract

1

$1$ from

- 1

$- 1$ .

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

$f (- 1) = | - 2 | + 4$

Step 3.1.2.1.2

The absolute value is the distance between a number and zero. The distance between

- 2

$- 2$ and

0

$0$ is

2

$2$ .

f (- 1) = 2 + 4

$f (- 1) = 2 + 4$

f (- 1) = 2 + 4

$f (- 1) = 2 + 4$

Step 3.1.2.2

Add

2

$2$ and

4

$4$ .

f (- 1) = 6

$f (- 1) = 6$

Step 3.1.2.3

The final answer is

6

$6$ .

y = 6

$y = 6$

y = 6

$y = 6$

y = 6

$y = 6$

Step 3.2

Substitute the

x

$x$ value

0

$0$ into

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

$f (x) = | x - 1 | + 4$ . In this case, the point is

(0, 5)

$(0, 5)$ .

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Step 3.2.1

Replace the variable

x

$x$ with

0

$0$ in the expression.

f (0) = | (0) - 1 | + 4

$f (0) = | (0) - 1 | + 4$

Step 3.2.2

Simplify the result.

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Step 3.2.2.1

Simplify each term.

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Step 3.2.2.1.1

Subtract

1

$1$ from

0

$0$ .

f (0) = | - 1 | + 4

$f (0) = | - 1 | + 4$

Step 3.2.2.1.2

The absolute value is the distance between a number and zero. The distance between

- 1

$- 1$ and

0

$0$ is

1

$1$ .

f (0) = 1 + 4

$f (0) = 1 + 4$

f (0) = 1 + 4

$f (0) = 1 + 4$

Step 3.2.2.2

Add

1

$1$ and

4

$4$ .

f (0) = 5

$f (0) = 5$

Step 3.2.2.3

The final answer is

5

$5$ .

y = 5

$y = 5$

y = 5

$y = 5$

y = 5

$y = 5$

Step 3.3

Substitute the

x

$x$ value

3

$3$ into

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

$f (x) = | x - 1 | + 4$ . In this case, the point is

(3, 6)

$(3, 6)$ .

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Step 3.3.1

Replace the variable

x

$x$ with

3

$3$ in the expression.

f (3) = | (3) - 1 | + 4

$f (3) = | (3) - 1 | + 4$

Step 3.3.2

Simplify the result.

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Step 3.3.2.1

Simplify each term.

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Step 3.3.2.1.1

Subtract

1

$1$ from

3

$3$ .

f (3) = | 2 | + 4

$f (3) = | 2 | + 4$

Step 3.3.2.1.2

The absolute value is the distance between a number and zero. The distance between

0

$0$ and

2

$2$ is

2

$2$ .

f (3) = 2 + 4

$f (3) = 2 + 4$

f (3) = 2 + 4

$f (3) = 2 + 4$

Step 3.3.2.2

Add

2

$2$ and

4

$4$ .

f (3) = 6

$f (3) = 6$

Step 3.3.2.3

The final answer is

6

$6$ .

y = 6

$y = 6$

y = 6

$y = 6$

y = 6

$y = 6$

Step 3.4

The absolute value can be graphed using the points around the vertex

(1, 4), (- 1, 6), (0, 5), (2, 5), (3, 6)

$(1, 4), (- 1, 6), (0, 5), (2, 5), (3, 6)$

\begin{array}{c} x & y \\ - 1 & 6 \\ 0 & 5 \\ 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{array}

$\begin{array}{c} x & y \\ - 1 & 6 \\ 0 & 5 \\ 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{array}$

\begin{array}{c} x & y \\ - 1 & 6 \\ 0 & 5 \\ 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{array}

$\begin{array}{c} x & y \\ - 1 & 6 \\ 0 & 5 \\ 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{array}$

Step 4