Graph F(x)=|3x+1|

$F (x) = | 3 x + 1 |$

Find the absolute value vertex. In this case, the vertex for $y = | 3 x + 1 |$ is $(- \frac{1}{3}, 0)$ .

To find the $x$ coordinate of the vertex, set the inside of the absolute value $3 x + 1$ equal to $0$ . In this case, $3 x + 1 = 0$ .

$3 x + 1 = 0$

Solve the equation $3 x + 1 = 0$ to find the $x$ coordinate for the absolute value vertex.

Subtract $1$ from both sides of the equation.

$3 x = - 1$

Divide each term in $3 x = - 1$ by $3$ and simplify.

Divide each term in $3 x = - 1$ by $3$ .

$\frac{3 x}{3} = \frac{- 1}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 1}{3}$

Divide $x$ by $1$ .

$x = \frac{- 1}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{3}$

Replace the variable $x$ with $- \frac{1}{3}$ in the expression.

$y = | 3 (- \frac{1}{3}) + 1 |$

Simplify $| 3 (- \frac{1}{3}) + 1 |$ .

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{1}{3}$ into the numerator.

$y = | 3 (\frac{- 1}{3}) + 1 |$

Cancel the common factor.

$y = | 3 (\frac{- 1}{3}) + 1 |$

Rewrite the expression.

$y = | - 1 + 1 |$

Add $- 1$ and $1$ .

$y = | 0 |$

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

$y = 0$

The absolute value vertex is $(- \frac{1}{3}, 0)$ .

$(- \frac{1}{3}, 0)$

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 $f (x) = | 3 x + 1 |$ . In this case, the point is $(- 2, 5)$ .

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

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

Simplify the result.

Multiply $3$ by $- 2$ .

$f (- 2) = | - 6 + 1 |$

Add $- 6$ and $1$ .

$f (- 2) = | - 5 |$

The absolute value is the distance between a number and zero. The distance between $- 5$ and $0$ is $5$ .

$f (- 2) = 5$

The final answer is $5$ .

$y = 5$

Substitute the $x$ value $- 1$ into $f (x) = | 3 x + 1 |$ . In this case, the point is $(- 1, 2)$ .

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

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

Simplify the result.

Multiply $3$ by $- 1$ .

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

Add $- 3$ and $1$ .

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

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

$f (- 1) = 2$

The final answer is $2$ .

$y = 2$

Substitute the $x$ value $0$ into $f (x) = | 3 x + 1 |$ . In this case, the point is $(0, 1)$ .

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

$f (0) = | 3 (0) + 1 |$

Simplify the result.

Multiply $3$ by $0$ .

$f (0) = | 0 + 1 |$

Add $0$ and $1$ .

$f (0) = | 1 |$

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

$f (0) = 1$

The final answer is $1$ .

$y = 1$

Substitute the $x$ value $1$ into $f (x) = | 3 x + 1 |$ . In this case, the point is $(1, 4)$ .

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

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

Simplify the result.

Multiply $3$ by $1$ .

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

Add $3$ and $1$ .

$f (1) = | 4 |$

The absolute value is the distance between a number and zero. The distance between $0$ and $4$ is $4$ .

$f (1) = 4$

The final answer is $4$ .

$y = 4$

The absolute value can be graphed using the points around the vertex $(- \frac{1}{3}, 0), (- 2, 5), (- 1, 2), (0, 1), (1, 4)$

$\begin{array}{c} x & y \\ - 2 & 5 \\ - 1 & 2 \\ - \frac{1}{3} & 0 \\ 0 & 1 \\ 1 & 4 \end{array}$

Graph F(x)=|3x+1|

Related Questions

Support LCMGCF.com by disabling your adblocker.