Question

What are the asymptotes and removable discontinuities, if any, of `f(x)= ((2x-3)(x+2))/(x-2)`?

Answer 1

`f(x) = ((2x-3)(x+2))/(x-2)`

Asymptotes: "Unreachable value that occurs when a denominator equals zero"

To find the value that makes our denominator equal to `0`, we set the component equal to `0` and solve for `x`:

`x-2 = 0`

`x=2`

So, when `x=2`, the denominator becomes zero. And, as we know, dividing by zero creates an asymptote; a value that infinitely approaches a point, but never reaches it

graph{y=((2x-3)(x+2))/(x-2)}

Notice how the line `x=2` is never reached, but becomes closer and closer

`color(white)(000)`

`color(white)(000)`
A "removable discontinuity," also known as a hole, occurs when a term in the numerator and denominator divides out

`color(white)(000)`
Since there are no terms that are the same in both the numerator and denominator, there are no terms that can divide out, thus, `color(green)(there)` `color(green)(are)` `color(green)(no)` `color(green)(ho l es)`