# How do you graph f(x)=4/(x-1)+1 using holes, vertical and horizontal asymptotes, x and y intercepts?

Jan 14, 2017

see explanation.

#### Explanation:

We can express f(x) as a single rational function.

$\Rightarrow f \left(x\right) = \frac{4}{x - 1} + \frac{x - 1}{x - 1} = \frac{x + 3}{x - 1}$

$\textcolor{b l u e}{\text{Asymptotes}}$

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 : $x - 1 = 0 \Rightarrow x = 1 \text{ is the asymptote}$

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by x

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

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

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

Holes occur when there is a duplicate factor on the numerator/denominator. This is not the case here, hence there are no holes.

$\textcolor{b l u e}{\text{Approaches to vertical asymptote}}$

${\lim}_{x \to {1}^{-}} = - \infty \text{ and } {\lim}_{x \to {1}^{+}} = + \infty$

$\textcolor{b l u e}{\text{Intercepts}}$

$x = 0 \to f \left(0\right) = \frac{3}{- 1} = - 3 \leftarrow \text{ y-intercept}$

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