Question

How do you graph and find the discontinuities of `y=(-4x-3)/(x-2)`?

Answer 1

`f(x) = -(4x+3)/(x-2)`
`= -(4x-8+8+3)/(x-2)`
`= -4-11/(x-2)`

A discontinuity at `x = a` occurs when:
1. `f(a)` does not exist
2. `lim_(x rarr a^+) f(x) ne lim_(x rarr a^-) f(x)`
3. `lim_(x rarr a) f(x) ne f(a)`

In this case, discontinuity at `x = 2` since `f(2)` does not exist.

To plot the graph:
`x`-axis intercept
`f(x) = 0`
`x = -3/4`

`y`-axis intercept
`f(0) = 3/2`

Vertical aysmptote at `x = 2`
`lim_(x rarr 2^(""_+^-)) f(x) = +- oo`

Horizontal asymptote at `y = -4`
`lim_(x rarr ""_+^(-)oo) f(x) = -4^(+-)`

No stationary points since there are no real values of `x` that satisfy `f'(x) = 0`.

graph{(-4x-3)/(x-2) [-5, 5, -20, 10]}