Graph y=|x+5|-1

$y = | x + 5 | - 1$

Step 1

Find the absolute value vertex. In this case, the vertex for $y = | x + 5 | - 1$ is $(- 5, - 1)$ .

To find the $x$ coordinate of the vertex, set the inside of the absolute value $x + 5$ equal to $0$ . In this case, $x + 5 = 0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

Replace the variable $x$ with $- 5$ in the expression.

$y = | (- 5) + 5 | - 1$

Simplify $| (- 5) + 5 | - 1$ .

Simplify each term.

Add $- 5$ and $5$ .

$y = | 0 | - 1$

The absolute value is the distance between a number and zero. The distance between $0$ and $0$ is $0$ .

$y = 0 - 1$

Subtract $1$ from $0$ .

$y = - 1$

The absolute value vertex is $(- 5, - 1)$ .

$(- 5, - 1)$

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:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Step 3

For each $x$ value, there is one $y$ value. Select a 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 $- 7$ into $f (x) = | x + 5 | - 1$ . In this case, the point is $(- 7, 1)$ .

Replace the variable $x$ with $- 7$ in the expression.

$f (- 7) = | (- 7) + 5 | - 1$

Simplify the result.

Simplify each term.

Add $- 7$ and $5$ .

$f (- 7) = | - 2 | - 1$

The absolute value is the distance between a number and zero. The distance between $- 2$ and $0$ is $2$ .

$f (- 7) = 2 - 1$

Subtract $1$ from $2$ .

$f (- 7) = 1$

The final answer is $1$ .

$y = 1$

Substitute the $x$ value $- 6$ into $f (x) = | x + 5 | - 1$ . In this case, the point is $(- 6, 0)$ .

Replace the variable $x$ with $- 6$ in the expression.

$f (- 6) = | (- 6) + 5 | - 1$

Simplify the result.

Simplify each term.

Add $- 6$ and $5$ .

$f (- 6) = | - 1 | - 1$

The absolute value is the distance between a number and zero. The distance between $- 1$ and $0$ is $1$ .

$f (- 6) = 1 - 1$

Subtract $1$ from $1$ .

$f (- 6) = 0$

The final answer is $0$ .

$y = 0$

Substitute the $x$ value $- 3$ into $f (x) = | x + 5 | - 1$ . In this case, the point is $(- 3, 1)$ .

Replace the variable $x$ with $- 3$ in the expression.

$f (- 3) = | (- 3) + 5 | - 1$

Simplify the result.

Simplify each term.

Add $- 3$ and $5$ .

$f (- 3) = | 2 | - 1$

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$f (- 3) = 2 - 1$

Subtract $1$ from $2$ .

$f (- 3) = 1$

The final answer is $1$ .

$y = 1$

The absolute value can be graphed using the points around the vertex $(- 5, - 1), (- 7, 1), (- 6, 0), (- 4, 0), (- 3, 1)$

$\begin{array}{c} x & y \\ - 7 & 1 \\ - 6 & 0 \\ - 5 & - 1 \\ - 4 & 0 \\ - 3 & 1 \end{array}$

Step 4

Graph y=|x+5|-1

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