Graph f^-1(x)=x/3+4/3

$f^{- 1} (x) = \frac{x}{3} + \frac{4}{3}$

Step 1

Rewrite the function as an equation.

$y = \frac{x}{3} + \frac{4}{3}$

Step 2

Rewrite in slope-intercept form.

The slope-intercept form is $y = m x + b$ , where $m$ is the slope and $b$ is the y-intercept.

$y = m x + b$

Reorder terms.

$y = \frac{1}{3} x + \frac{4}{3}$

Step 3

Use the slope-intercept form to find the slope and y-intercept.

Find the values of $m$ and $b$ using the form $y = m x + b$ .

$m = \frac{1}{3}$

$b = \frac{4}{3}$

The slope of the line is the value of $m$ , and the y-intercept is the value of $b$ .

Slope: $\frac{1}{3}$

y-intercept: $(0, \frac{4}{3})$

Slope: $\frac{1}{3}$

y-intercept: $(0, \frac{4}{3})$

Step 4

Any line can be graphed using two points. Select two $x$ values, and plug them into the equation to find the corresponding $y$ values.

Reorder terms.

$y = \frac{1}{3} x + \frac{4}{3}$

Find the x-intercept.

To find the x-intercept(s), substitute in $0$ for $y$ and solve for $x$ .

$0 = \frac{1}{3} x + \frac{4}{3}$

Solve the equation.

Rewrite the equation as $\frac{1}{3} x + \frac{4}{3} = 0$ .

$\frac{1}{3} x + \frac{4}{3} = 0$

Combine $\frac{1}{3}$ and $x$ .

$\frac{x}{3} + \frac{4}{3} = 0$

Subtract $\frac{4}{3}$ from both sides of the equation.

$\frac{x}{3} = - \frac{4}{3}$

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

$x = - 4$

x-intercept(s) in point form.

x-intercept(s): $(- 4, 0)$

Find the y-intercept.

To find the y-intercept(s), substitute in $0$ for $x$ and solve for $y$ .

$y = \frac{1}{3} \cdot (0) + \frac{4}{3}$

Solve the equation.

Multiply $\frac{1}{3}$ by $0$ .

$y = \frac{1}{3} \cdot 0 + \frac{4}{3}$

Remove parentheses.

$y = \frac{1}{3} \cdot (0) + \frac{4}{3}$

Simplify $\frac{1}{3} \cdot (0) + \frac{4}{3}$ .

Multiply $\frac{1}{3}$ by $0$ .

$y = 0 + \frac{4}{3}$

Add $0$ and $\frac{4}{3}$ .

$y = \frac{4}{3}$

y-intercept(s) in point form.

y-intercept(s): $(0, \frac{4}{3})$

Create a table of the $x$ and $y$ values.

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

Step 5

Graph the line using the slope and the y-intercept, or the points.

Slope: $\frac{1}{3}$

y-intercept: $(0, \frac{4}{3})$

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

Step 6

Graph f^-1(x)=x/3+4/3

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