# How do you graph using the intercepts for 5x + 2y = -2?

Nov 9, 2015

Plot the intercept points and draw a straight line through them
graph{(5x+2y+2) ((x+2/5)^2+y^2-0.01)(x^2+(y+1)^2-0.01)=0 [-4.93, 4.934, -2.465, 2.465]}

#### Explanation:

The x and y intercepts give us two points on the line defined by the linear equation.

The x-intercept is the value of $x$ when $y = 0$ (for all points on the X-axis $y = 0$)

Given $5 x + 2 y = - 2$
when $y = 0$
$\Rightarrow \textcolor{w h i t e}{\text{XXX}} x = - \frac{2}{5}$
which gives the point $\left(- \frac{2}{5} , 0\right)$

Similarly for the y-intercept
we get the point $\left(0 , - 1\right)$