Graph y=e^(|x|)

$y = e^{| x |}$

Find the absolute value vertex. In this case, the vertex for $y = e^{| x |}$ is $(0, 1)$ .

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

$x = 0$

Replace the variable $x$ with $0$ in the expression.

$y = e^{| 0 |}$

Simplify $e^{| 0 |}$ .

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

$y = e^{0}$

Anything raised to $0$ is $1$ .

$y = 1$

The absolute value vertex is $(0, 1)$ .

$(0, 1)$

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 $- 2$ into $f (x) = e^{| x |}$ . In this case, the point is $(- 2, e^{2})$ .

Replace the variable $x$ with $- 2$ in the expression.

$f (- 2) = e^{| - 2 |}$

Simplify the result.

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

$f (- 2) = e^{2}$

The final answer is $e^{2}$ .

$y = e^{2}$

Substitute the $x$ value $- 1$ into $f (x) = e^{| x |}$ . In this case, the point is $(- 1, e)$ .

Replace the variable $x$ with $- 1$ in the expression.

$f (- 1) = e^{| - 1 |}$

Simplify the result.

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

$f (- 1) = e$

The final answer is $e$ .

$y = e$

Substitute the $x$ value $2$ into $f (x) = e^{| x |}$ . In this case, the point is $(2, e^{2})$ .

Replace the variable $x$ with $2$ in the expression.

$f (2) = e^{| 2 |}$

Simplify the result.

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

$f (2) = e^{2}$

The final answer is $e^{2}$ .

$y = e^{2}$

The absolute value can be graphed using the points around the vertex $(0, 1), (- 2, 7.39), (- 1, 2.72), (1, 2.72), (2, 7.39)$

$\begin{array}{c} x & y \\ - 2 & 7.39 \\ - 1 & 2.72 \\ 0 & 1 \\ 1 & 2.72 \\ 2 & 7.39 \end{array}$

Graph y=e^(|x|)

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