# How do you graph y=(x+2)/(x+3) using asymptotes, intercepts, end behavior?

Dec 20, 2016

graph{(x+2)/(x+3) [-10, 10, -5, 5]}

$y$-intercept is $\frac{2}{3}$

$x$-intercept is $- 2$

vertical asymptote at $x = - 3$

$y < 1$ when $x > - 3$

$y > 1$ when $x < - 3$

#### Explanation:

asymptote:

$\frac{n}{0}$ = undefined

$\therefore$ if $x + 3 = 0$, $y$ is undefined.

this means that it is not on the graph, and so is shown as the asymptote.

when $x + 3 = 0$, $x = 0 - 3$

$x = - 3$

intercepts:

$y$:

the $y$-intercept is when $x = 0$

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

$y = \frac{2}{3}$

$y$-intercept is $\frac{2}{3}$

$x$:

the $x$-intercept is when $y = 0$

$\frac{x + 2}{x + 3} = 0$

the numerator has to be $0$, since $\frac{0}{n} = 0$

this means that $x + 2 = 0$

when $x + 2 = 0$, $x = 0 - 2$

$x$-intercept is $- 2$

end behaviour:

$y < 1$

$x + 2$ is always less than $x + 3$. with both positive and negative numbers.

the fraction $\frac{x + 2}{x + 3}$ cannot be simplified to $\frac{\ge 1}{1}$ if $x > - 3$

however, it is vice versa for negative numbers, since smaller negative numbers have a higher absolute value (distance from $0$),

e.g. $- \frac{3}{-} 2 = | 3 \frac{|}{|} 2 | = \frac{3}{2}$

this means that $\frac{x + 2}{x + 3}$ cannot be simplified to $\frac{\le 1}{1}$ if $x < - 3$