Graph f(x)=3x+3|x|

$f (x) = 3 x + 3 | x |$

Find the absolute value vertex. In this case, the vertex for $y = 3 x + 3 | x |$ is $(0, 0)$ .

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.

$y = 3 (0) + 3 | 0 |$

Simplify $3 (0) + 3 | 0 |$ .

Simplify each term.

Multiply $3$ by $0$ .

$y = 0 + 3 | 0 |$

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

$y = 0 + 3 \cdot 0$

Multiply $3$ by $0$ .

$y = 0 + 0$

Add $0$ and $0$ .

$y = 0$

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

$(0, 0)$

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 $f (x) = 3 x + 3 | x |$ . In this case, the point is $(- 2, 0)$ .

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

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

Simplify the result.

Simplify each term.

Multiply $3$ by $- 2$ .

$f (- 2) = - 6 + 3 | - 2 |$

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

$f (- 2) = - 6 + 3 \cdot 2$

Multiply $3$ by $2$ .

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

Add $- 6$ and $6$ .

$f (- 2) = 0$

The final answer is $0$ .

$y = 0$

Substitute the $x$ value $1$ into $f (x) = 3 x + 3 | x |$ . In this case, the point is $(1, 6)$ .

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

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

Simplify the result.

Simplify each term.

Multiply $3$ by $1$ .

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

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

$f (1) = 3 + 3 \cdot 1$

Multiply $3$ by $1$ .

$f (1) = 3 + 3$

Add $3$ and $3$ .

$f (1) = 6$

The final answer is $6$ .

$y = 6$

Substitute the $x$ value $2$ into $f (x) = 3 x + 3 | x |$ . In this case, the point is $(2, 12)$ .

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

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

Simplify the result.

Simplify each term.

Multiply $3$ by $2$ .

$f (2) = 6 + 3 | 2 |$

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

$f (2) = 6 + 3 \cdot 2$

Multiply $3$ by $2$ .

$f (2) = 6 + 6$

Add $6$ and $6$ .

$f (2) = 12$

The final answer is $12$ .

$y = 12$

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

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

Graph f(x)=3x+3|x|

Related Questions

Support LCMGCF.com by disabling your adblocker.