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

Jan 3, 2018

$\text{see explanation}$

#### Explanation:

$\text{factorise the numerator/denominator}$

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

$\text{the cancelling of the factor x indicates a hole at x = 0}$

$\text{substitute x = 0 into } f \left(x\right) = \frac{x + 3}{x - 1}$

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

$\Rightarrow \text{there is a hole at } \left(0 , - 3\right)$

$\text{the graph of "(x+3)/(x-1)" is the same as the graph of}$

$\frac{{x}^{2} + 3 x}{{x}^{2} - x} \text{ but without the hole}$

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.

$\text{solve "x-1=0rArrx=1" is the asymptote}$

$\text{horizontal asymptotes occur as}$

${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ (a constant)}$

$\text{divide terms on numerator/denominator by x}$

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

$\text{as } x \to \pm \infty , f \left(x\right) \to \frac{1 + 0}{1 - 0}$

$\Rightarrow y = 1 \text{ is the asymptote}$

$\textcolor{b l u e}{\text{for intercepts}}$

• " let x = 0 for y-intercept"

• " let y = 0 for x-intercepts"

$f \left(0\right) = - 3 \leftarrow \textcolor{red}{\text{y-intercept}}$

$y = 0 \to x + 3 = 0 \Rightarrow x = - 3 \leftarrow \textcolor{red}{\text{x-intercept}}$
graph{(x+3)/(x-1) [-10, 10, -5, 5]}