# How do you graph f(x)=(3x+8)/(x−2) using holes, vertical and horizontal asymptotes, x and y intercepts?

How do you graph $f\left(x\right)=\frac{3x+8}{x-2}$ using holes, vertical and horizontal asymptotes, x and y intercepts?
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Jamari Morgan
Asymptotes
The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.
solve : $x-2=0⇒x=2\phantom{\rule{1ex}{0ex}}\text{is the asymptote}$
Horizontal asymptotes occur as
$\underset{x\to ±\infty }{lim},f\left(x\right)\to c\phantom{\rule{1ex}{0ex}}\text{( a constant)}$
divide terms on numerator/denominator by x
$f\left(x\right)=\frac{\frac{3x}{x}+\frac{8}{x}}{\frac{x}{x}-\frac{2}{x}}=\frac{3+\frac{8}{x}}{1-\frac{2}{x}}$
as $x\to ±\infty ,f\left(x\right)\to \frac{3+0}{1-0}$
$⇒y=3\phantom{\rule{1ex}{0ex}}\text{is the asymptote}$
Holes occur when there is a duplicate factor on the numerator/denominator. This is not the case here, hence there are no holes.
Intercepts
$x=0\to f\left(0\right)=\frac{8}{-2}=-4←\phantom{\rule{1ex}{0ex}}\text{y-intercept}$
$y=0⇒3x+8=0⇒x=-\frac{8}{3}←\phantom{\rule{1ex}{0ex}}\text{x-intercept}$
graph{(3x+8)/(x-2) [-20, 20, -10, 10]}