# How do you graph using the intercepts for y=-6x-9?

Mar 11, 2018

See a solution process below:

#### Explanation:

Because this equation is in slope-intercept form we can find the $x$-intercept and $y$-intercept directly from the equation. The slope-intercept form of a linear equation is: $y = \textcolor{red}{m} x + \textcolor{b l u e}{b}$

Where $\textcolor{red}{m}$ is the slope, $\textcolor{b l u e}{b}$ is the y-intercept value and $\frac{\textcolor{b l u e}{- b}}{\textcolor{red}{m}}$ is the x-intercept..

$y = \textcolor{red}{- 6} x - \textcolor{b l u e}{9}$

Therefore the $y$-intercept is: $\textcolor{b l u e}{b = - 9}$ or $\left(0 , \textcolor{b l u e}{- 9}\right)$

And, the $x$-intercept is:

$\frac{\textcolor{b l u e}{- - 9}}{\textcolor{red}{- 6}} \implies \frac{\textcolor{b l u e}{9}}{\textcolor{red}{- 6}} \implies - \frac{\textcolor{b l u e}{3 \times 3}}{\textcolor{red}{3 \times 2}} \implies - \frac{\textcolor{b l u e}{\textcolor{b l a c k}{\cancel{\textcolor{b l u e}{3}}} \times 3}}{\textcolor{red}{\textcolor{b l a c k}{\cancel{\textcolor{red}{3}}} \times 2}} \implies - \frac{3}{2}$

Or

#(-3/2, 0)

We can next plot the two points on the coordinate plane:

graph{(x^2+(y+9)^2-0.3)((x+ 3/2)^2+y^2-0.3)=0 [-30, 30, -15, 15]}

Now, we can draw a straight line through the two points to graph the line:

graph{(y+6x+9)(x^2+(y+9)^2-0.3)((x+ 3/2)^2+y^2-0.3)=0 [-30, 30, -15, 15]}