Graph y(x)=|5x-3|+4

$y (x) = | 5 x - 3 | + 4$

Step 1

Find the absolute value vertex. In this case, the vertex for $y = | 5 x - 3 | + 4$ is $(\frac{3}{5}, 4)$ .

To find the $x$ coordinate of the vertex, set the inside of the absolute value $5 x - 3$ equal to $0$ . In this case, $5 x - 3 = 0$ .

$5 x - 3 = 0$

Solve the equation $5 x - 3 = 0$ to find the $x$ coordinate for the absolute value vertex.

Add $3$ to both sides of the equation.

$5 x = 3$

Divide each term in $5 x = 3$ by $5$ and simplify.

Divide each term in $5 x = 3$ by $5$ .

$\frac{5 x}{5} = \frac{3}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{3}{5}$

Divide $x$ by $1$ .

$x = \frac{3}{5}$

Replace the variable $x$ with $\frac{3}{5}$ in the expression.

$y = | 5 (\frac{3}{5}) - 3 | + 4$

Simplify $| 5 (\frac{3}{5}) - 3 | + 4$ .

Simplify each term.

Cancel the common factor of $5$ .

Cancel the common factor.

$y = | 5 (\frac{3}{5}) - 3 | + 4$

Rewrite the expression.

$y = | 3 - 3 | + 4$

Subtract $3$ from $3$ .

$y = | 0 | + 4$

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

$y = 0 + 4$

Add $0$ and $4$ .

$y = 4$

The absolute value vertex is $(\frac{3}{5}, 4)$ .

$(\frac{3}{5}, 4)$

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 $- 1$ into $f (x) = | 5 x - 3 | + 4$ . In this case, the point is $(- 1, 12)$ .

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

$f (- 1) = | 5 (- 1) - 3 | + 4$

Simplify the result.

Simplify each term.

Multiply $5$ by $- 1$ .

$f (- 1) = | - 5 - 3 | + 4$

Subtract $3$ from $- 5$ .

$f (- 1) = | - 8 | + 4$

The absolute value is the distance between a number and zero. The distance between $- 8$ and $0$ is $8$ .

$f (- 1) = 8 + 4$

Add $8$ and $4$ .

$f (- 1) = 12$

The final answer is $12$ .

$y = 12$

Substitute the $x$ value $0$ into $f (x) = | 5 x - 3 | + 4$ . In this case, the point is $(0, 7)$ .

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

$f (0) = | 5 (0) - 3 | + 4$

Simplify the result.

Simplify each term.

Multiply $5$ by $0$ .

$f (0) = | 0 - 3 | + 4$

Subtract $3$ from $0$ .

$f (0) = | - 3 | + 4$

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

$f (0) = 3 + 4$

Add $3$ and $4$ .

$f (0) = 7$

The final answer is $7$ .

$y = 7$

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

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

$f (1) = | 5 (1) - 3 | + 4$

Simplify the result.

Simplify each term.

Multiply $5$ by $1$ .

$f (1) = | 5 - 3 | + 4$

Subtract $3$ from $5$ .

$f (1) = | 2 | + 4$

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

$f (1) = 2 + 4$

Add $2$ and $4$ .

$f (1) = 6$

The final answer is $6$ .

$y = 6$

Substitute the $x$ value $2$ into $f (x) = | 5 x - 3 | + 4$ . In this case, the point is $(2, 11)$ .

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

$f (2) = | 5 (2) - 3 | + 4$

Simplify the result.

Simplify each term.

Multiply $5$ by $2$ .

$f (2) = | 10 - 3 | + 4$

Subtract $3$ from $10$ .

$f (2) = | 7 | + 4$

The absolute value is the distance between a number and zero. The distance between $0$ and $7$ is $7$ .

$f (2) = 7 + 4$

Add $7$ and $4$ .

$f (2) = 11$

The final answer is $11$ .

$y = 11$

The absolute value can be graphed using the points around the vertex $(\frac{3}{5}, 4), (- 1, 12), (0, 7), (1, 6), (2, 11)$

$\begin{array}{c} x & y \\ - 1 & 12 \\ 0 & 7 \\ \frac{3}{5} & 4 \\ 1 & 6 \\ 2 & 11 \end{array}$

Step 4

Graph y(x)=|5x-3|+4

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