Graph f(x)>=-|x-5|-3

$f (x) \geq - | x - 5 | - 3$

Find the absolute value vertex. In this case, the vertex for $y = - | x - 5 | - 3$ is $(5, - 3)$ .

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

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

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

$y = - | (5) - 5 | - 3$

Simplify $- | (5) - 5 | - 3$ .

Simplify each term.

Subtract $5$ from $5$ .

$y = - | 0 | - 3$

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

$y = - 0 - 3$

Multiply $- 1$ by $0$ .

$y = 0 - 3$

Subtract $3$ from $0$ .

$y = - 3$

The absolute value vertex is $(5, - 3)$ .

$(5, - 3)$

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:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

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

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

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

Simplify the result.

Simplify each term.

Subtract $5$ from $3$ .

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

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

$f (3) = - 1 \cdot 2 - 3$

Multiply $- 1$ by $2$ .

$f (3) = - 2 - 3$

Subtract $3$ from $- 2$ .

$f (3) = - 5$

The final answer is $- 5$ .

$y = - 5$

Substitute the $x$ value $4$ into $f (x) = - | x - 5 | - 3$ . In this case, the point is $(4, - 4)$ .

Replace the variable $x$ with $4$ in the expression.

$f (4) = - | (4) - 5 | - 3$

Simplify the result.

Simplify each term.

Subtract $5$ from $4$ .

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

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

$f (4) = - 1 \cdot 1 - 3$

Multiply $- 1$ by $1$ .

$f (4) = - 1 - 3$

Subtract $3$ from $- 1$ .

$f (4) = - 4$

The final answer is $- 4$ .

$y = - 4$

Substitute the $x$ value $7$ into $f (x) = - | x - 5 | - 3$ . In this case, the point is $(7, - 5)$ .

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

$f (7) = - | (7) - 5 | - 3$

Simplify the result.

Simplify each term.

Subtract $5$ from $7$ .

$f (7) = - | 2 | - 3$

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

$f (7) = - 1 \cdot 2 - 3$

Multiply $- 1$ by $2$ .

$f (7) = - 2 - 3$

Subtract $3$ from $- 2$ .

$f (7) = - 5$

The final answer is $- 5$ .

$y = - 5$

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

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

Graph f(x)>=-|x-5|-3

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