# How do you graph f(x)=(x+2)(x-1)/(x-3)?

Jul 24, 2018

#### Explanation:

Given: $f \left(x\right) = \left(x + 2\right) \frac{\left(x - 1\right)}{x - 3}$

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

This type of equation is called a rational (fraction) function.

It has the form: $f \left(x\right) = \frac{N \left(x\right)}{D \left(x\right)} = \frac{{a}_{n} {x}^{n} + \ldots}{{b}_{m} {x}^{m} + \ldots}$,

where N(x)) is the numerator and $D \left(x\right)$ is the denominator,

$n$ = the degree of $N \left(x\right)$ and $m$ = the degree of $\left(D \left(x\right)\right)$

and ${a}_{n}$ is the leading coefficient of the $N \left(x\right)$ and

${b}_{m}$ is the leading coefficient of the $D \left(x\right)$

Step 1 factor : The given function is already factored.

Step 2, cancel any factors that are both in $\left(N \left(x\right)\right)$ and D(x)) (determines holes):

The given function has no holes $\text{ "=> " no factors that cancel}$

Step 3, find vertical asymptotes: $D \left(x\right) = 0$

vertical asymptote at $x = 3$

Step 4, find horizontal asymptotes:
Compare the degrees:

If $n < m$ the horizontal asymptote is $y = 0$

If $n = m$ the horizontal asymptote is $y = {a}_{n} / {b}_{m}$

If $n > m$ there are no horizontal asymptotes

In the given equation: n = 2; m =1

There is no horizontal asymptotes

Step 5, find slant or oblique asymptote when $n = m + 1$:

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

Use Long division:

" "ul(" "color(red)(x + 4) +10/(x-3))
$x - 3 | {x}^{2} + x - 2$
$\text{ } \underline{{x}^{2} - 3 x}$
$\text{ } 4 x - 2$
$\text{ } \underline{4 x - 12}$
$\text{ } 10$

slant or oblique asymptote: $\textcolor{red}{y = x + 4}$

Step 6, find x-intercept(s) : $N \left(x\right) = 0$

$\left(x + 2\right) \left(x - 1\right) = 0$
$\text{ "x + 2 = 0 " "=> x"-intercept} \left(- 2 , 0\right)$
$\text{ "x -1 = 0 " "=> x"-intercept} \left(1 , 0\right)$

Step 7, find y-intercept(s): $x = 0$

$f \left(0\right) = \left(0 + 2\right) \frac{\left(0 - 1\right)}{0 - 3} = 2 \frac{- 1}{-} 3 = \frac{2}{3}$

$y$-intercept: (0, 2/3)#

Graph of $f \left(x\right) = \left(x + 2\right) \frac{\left(x - 1\right)}{x - 3}$:

graph{(x+2)((x-1))/(x-3) [-50, 50, -50, 50]}