Question

construct a rational function that has a vertical asymptote at x = 3 and a removable discontinuity at x = -2.

Answer 1

Here is a clever way to do it.

You want a discontinuity at x=3, so you should have (x-3) showing up somewhere in the function. We need another discontinuity at x=-2, so you should have (x+2) somewhere.

When I say "somewhere" I actually mean denominator of the rational function. That's what produces these discontinuities, the denominator.\[\large\rm f(x)=\frac{}{(x-3)(x+2)}\]What should we put in the numerator though?

Well, we want this discontinuity at x=-2 to be `removable`.
So the factor in the denominator, (x+2), should cancel out to give us this removed value.\[\large\rm f(x)=\frac{(x+2)}{(x-3)(x+2)}\]

Answer 2

So that's uhh.. ya.. that's one way to do it.

Answer 3

okay, makes sense. it doesn't matter if (x+2) is seen twice? due to the fact it's on the numerator and denominator?

Answer 4

It `must` show up twice.
That's what makes it a removable discontinuity.
It's sort of a special thing.

Usually when you have a removable discontinuity,
the factor will be well hidden.

Here is an example,\[\large\rm \frac{x^2+3x+2}{x^2+7x+10}\]In this example we have a removable discontinuity at x=-2 but it might not be so obvious at first glance. If we factor the numerator and denominator,\[\large\rm \frac{(x+1)(x+2)}{(x+5)(x+2)}\]you can see both the numerator and denominator have the (x+2), but it was well hidden, ya?

Answer 5

ohhh i see now, thank you for the explanation!

Answer 6

np