Question

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

Answer 1

The vertical asymptote is `x=7`
The horizontal asymptote is `y=0`
There are no oblique asymptotes.
The intercepts are `(0,20/7)` and `(20/3,0)`

Explanation:

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

`lim_(x->-oo)y=lim_(x->-oo)1/x=0^(-)`

`lim_(x->+oo)y=lim_(x->+oo)1/x=0^(+)`

Therefore `y=0` is a horizontal asymptote.

`lim_(x->7^-)y=-oo`

`lim_(x->7^+)y=+oo`

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

Intercepts, when `x=0`, `y=-1/7+3=20/7`

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