Graph f(x)>=-|3x+6|+1

$f (x) \geq - | 3 x + 6 | + 1$

Step 1

Find the absolute value vertex. In this case, the vertex for $y = - | 3 x + 6 | + 1$ is $(- 2, 1)$ .

To find the $x$ coordinate of the vertex, set the inside of the absolute value $3 x + 6$ equal to $0$ . In this case, $3 x + 6 = 0$ .

$3 x + 6 = 0$

Solve the equation $3 x + 6 = 0$ to find the $x$ coordinate for the absolute value vertex.

Subtract $6$ from both sides of the equation.

$3 x = - 6$

Divide each term in $3 x = - 6$ by $3$ and simplify.

Divide each term in $3 x = - 6$ by $3$ .

$\frac{3 x}{3} = \frac{- 6}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 6}{3}$

Divide $x$ by $1$ .

$x = \frac{- 6}{3}$

Simplify the right side.

Divide $- 6$ by $3$ .

$x = - 2$

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

$y = - | 3 (- 2) + 6 | + 1$

Simplify $- | 3 (- 2) + 6 | + 1$ .

Simplify each term.

Multiply $3$ by $- 2$ .

$y = - | - 6 + 6 | + 1$

Add $- 6$ and $6$ .

$y = - | 0 | + 1$

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

$y = - 0 + 1$

Multiply $- 1$ by $0$ .

$y = 0 + 1$

Add $0$ and $1$ .

$y = 1$

The absolute value vertex is $(- 2, 1)$ .

$(- 2, 1)$

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 $- 4$ into $f (x) = - | 3 x + 6 | + 1$ . In this case, the point is $(- 4, - 5)$ .

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

$f (- 4) = - | 3 (- 4) + 6 | + 1$

Simplify the result.

Simplify each term.

Multiply $3$ by $- 4$ .

$f (- 4) = - | - 12 + 6 | + 1$

Add $- 12$ and $6$ .

$f (- 4) = - | - 6 | + 1$

The absolute value is the distance between a number and zero. The distance between $- 6$ and $0$ is $6$ .

$f (- 4) = - 1 \cdot 6 + 1$

Multiply $- 1$ by $6$ .

$f (- 4) = - 6 + 1$

Add $- 6$ and $1$ .

$f (- 4) = - 5$

The final answer is $- 5$ .

$y = - 5$

Substitute the $x$ value $- 3$ into $f (x) = - | 3 x + 6 | + 1$ . In this case, the point is $(- 3, - 2)$ .

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

$f (- 3) = - | 3 (- 3) + 6 | + 1$

Simplify the result.

Simplify each term.

Multiply $3$ by $- 3$ .

$f (- 3) = - | - 9 + 6 | + 1$

Add $- 9$ and $6$ .

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

The absolute value is the distance between a number and zero. The distance between $- 3$ and $0$ is $3$ .

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

Multiply $- 1$ by $3$ .

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

Add $- 3$ and $1$ .

$f (- 3) = - 2$

The final answer is $- 2$ .

$y = - 2$

Substitute the $x$ value $0$ into $f (x) = - | 3 x + 6 | + 1$ . In this case, the point is $(0, - 5)$ .

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

$f (0) = - | 3 (0) + 6 | + 1$

Simplify the result.

Simplify each term.

Multiply $3$ by $0$ .

$f (0) = - | 0 + 6 | + 1$

Add $0$ and $6$ .

$f (0) = - | 6 | + 1$

The absolute value is the distance between a number and zero. The distance between $0$ and $6$ is $6$ .

$f (0) = - 1 \cdot 6 + 1$

Multiply $- 1$ by $6$ .

$f (0) = - 6 + 1$

Add $- 6$ and $1$ .

$f (0) = - 5$

The final answer is $- 5$ .

$y = - 5$

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

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

Step 4

Graph f(x)>=-|3x+6|+1

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