# How do you find the x and y intercept given f(x) = 2/3x + 4?

x-intercept: $= - 6$

y-intercept: $= 4$

#### Explanation:

The given equation of straight line is

$f \left(x\right) = \frac{2}{3} x + 4$

$y = \frac{2}{3} x + 4$

$- \frac{2}{3} x + y = 4$

$\frac{1}{4} \left(- \frac{2}{3} x + y\right) = 1$

$- \frac{1}{6} x + \frac{y}{4} = 1$

$\frac{x}{- 6} + \frac{y}{4} = 1$

Above equation is in standard intercept form: $\frac{x}{a} + \frac{y}{b} = 1$

which has

x-intercept: $a = - 6$

y-intercept: $b = 4$

Jul 28, 2018

The intercepts are $\left(- 6 , 0\right)$ on the $x$-axis
and $\left(0 , 4\right)$ on the $y$-axis.

#### Explanation:

$f \left(x\right) = \frac{2}{3} x + 4$ can be written as $y = \frac{2}{3} x + 4$

This is in slope-intercept form, $y = m x + c$ so we immediately know that the $y$ -intercept is $4$, the point $\left(0 , 4\right)$
On the $y$ -axis the $x$ -value is always $0$.

To find the $x$- intercept, set $y = 0$ because on the $x$-axis the $y$-value is always $0$

$0 = \frac{2}{3} x + 4 \text{ } \leftarrow \times 3$

$0 = 2 x + 12$

$- 12 = 2 x$

$- 6 = x \text{ } \leftarrow$ this is the $x$-intercept

The two intercepts are $\left(- 6 , 0\right) \mathmr{and} \left(0 , 4\right)$

We could also have changed the original equation to :

$3 y = 2 x + 12 \text{ }$ which is re-arranged to:

$2 x - 3 y = - 12 \text{ } \leftarrow$ this is standard form.

Set $x = 0$ to get $y = 4 \text{ } \leftarrow$ the $y$-intercept

Set $y = 0$ to get $x = - 6 \text{ } \leftarrow$ the $x$-intercept