Graph f(x)=x^2+5|x|+6

$f (x) = x^{2} + 5 | x | + 6$

Step 1

Find the absolute value vertex. In this case, the vertex for $y = x^{2} + 5 | x | + 6$ is $(0, 6)$ .

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.

$y = {(0)}^{2} + 5 | 0 | + 6$

Simplify ${(0)}^{2} + 5 | 0 | + 6$ .

Simplify each term.

Raising $0$ to any positive power yields $0$ .

$y = 0 + 5 | 0 | + 6$

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

$y = 0 + 5 \cdot 0 + 6$

Multiply $5$ by $0$ .

$y = 0 + 0 + 6$

Simplify by adding numbers.

Add $0$ and $0$ .

$y = 0 + 6$

Add $0$ and $6$ .

$y = 6$

The absolute value vertex is $(0, 6)$ .

$(0, 6)$

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 $- 2$ into $f (x) = x^{2} + 5 | x | + 6$ . In this case, the point is $(- 2, 20)$ .

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

$f (- 2) = {(- 2)}^{2} + 5 | - 2 | + 6$

Simplify the result.

Simplify each term.

Raise $- 2$ to the power of $2$ .

$f (- 2) = 4 + 5 | - 2 | + 6$

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

$f (- 2) = 4 + 5 \cdot 2 + 6$

Multiply $5$ by $2$ .

$f (- 2) = 4 + 10 + 6$

Simplify by adding numbers.

Add $4$ and $10$ .

$f (- 2) = 14 + 6$

Add $14$ and $6$ .

$f (- 2) = 20$

The final answer is $20$ .

$y = 20$

Substitute the $x$ value $- 1$ into $f (x) = x^{2} + 5 | x | + 6$ . In this case, the point is $(- 1, 12)$ .

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

$f (- 1) = {(- 1)}^{2} + 5 | - 1 | + 6$

Simplify the result.

Simplify each term.

Raise $- 1$ to the power of $2$ .

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

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

$f (- 1) = 1 + 5 \cdot 1 + 6$

Multiply $5$ by $1$ .

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

Simplify by adding numbers.

Add $1$ and $5$ .

$f (- 1) = 6 + 6$

Add $6$ and $6$ .

$f (- 1) = 12$

The final answer is $12$ .

$y = 12$

Substitute the $x$ value $2$ into $f (x) = x^{2} + 5 | x | + 6$ . In this case, the point is $(2, 20)$ .

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

$f (2) = {(2)}^{2} + 5 | 2 | + 6$

Simplify the result.

Simplify each term.

Raise $2$ to the power of $2$ .

$f (2) = 4 + 5 | 2 | + 6$

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

$f (2) = 4 + 5 \cdot 2 + 6$

Multiply $5$ by $2$ .

$f (2) = 4 + 10 + 6$

Simplify by adding numbers.

Add $4$ and $10$ .

$f (2) = 14 + 6$

Add $14$ and $6$ .

$f (2) = 20$

The final answer is $20$ .

$y = 20$

The absolute value can be graphed using the points around the vertex $(0, 6), (- 2, 20), (- 1, 12), (1, 12), (2, 20)$

$\begin{array}{c} x & y \\ - 2 & 20 \\ - 1 & 12 \\ 0 & 6 \\ 1 & 12 \\ 2 & 20 \end{array}$

Step 4

Graph f(x)=x^2+5|x|+6

Related Questions

Support LCMGCF.com by disabling your adblocker.