# How do you graph y=1/(x-7)+3 using asymptotes, intercepts, end behavior?

Nov 12, 2016

The vertical asymptote is $x = 7$
The horizontal asymptote is $y = 0$
There are no oblique asymptotes.
The intercepts are $\left(0 , \frac{20}{7}\right)$ and $\left(\frac{20}{3} , 0\right)$

#### Explanation:

As you cannot divide by $0$, a vertical asymptote is $x = 7$

${\lim}_{x \to - \infty} y = {\lim}_{x \to - \infty} \frac{1}{x} = {0}^{-}$

${\lim}_{x \to + \infty} y = {\lim}_{x \to + \infty} \frac{1}{x} = {0}^{+}$

Therefore $y = 0$ is a horizontal asymptote.

${\lim}_{x \to {7}^{-}} y = - \infty$

${\lim}_{x \to {7}^{+}} y = + \infty$

There are no oblique intercept as the degree of the numerator is $<$ the degree of the denominator.

Intercepts, when $x = 0$, $y = - \frac{1}{7} + 3 = \frac{20}{7}$

When $y = 0$ $\implies$$\frac{1}{x - 7} + 3 = 0$ $\implies$$x = \frac{20}{3}$
graph{3+1/(x-7) [-14.87, 25.66, -8.18, 12.09]}