Question

How do you identify all asymptotes or holes and intercepts for `f(x)=(x+3)/(x^2+7x+12)`?

Answer 1

First you need to factor the denominator.

`x^2+7x+12` will be factored into `( x+3)` and `(x+4)`

Now you will have `(x+3)/ ((x+3)(x+4))`

Now cancel out any like terms, in this case the `( x+3)` values.

Now `x= -3` becomes your x coordinate of the hole

to find your y component, plug in `x= -3` back into the remaining equation , which will be `1/ ( x+4)`

so `1/ ( -3 + 4 )`

This equals to `1/1` which just equals to 1

so the coordinate for your hole would be `( -3 , 1 )`

Now you still have the `( x + 4 )` remaining so `x= -4` would be your V.A.

• note that when two like values cancel out there is a whole and when there are no values cancelling out then the zero of `x` in the denominator will be your V.A. *

Now we can check for horizontal asymptotes. H.A.s exist only when the numerator's leading degree is less than or equal to the leading degree of the denominator.

In this case the degree of the numerator is x and the degree of the denominator is `x^2` thus an H.A exists. To find the H.A we just divide the degree of the numerator by the degree of the denominator. This looks like `( x ) / ( x^2 )`. If the degree of the denominator is greater then the H.A will always be `y= 0`. So we have an H.A. at `y= 0`.

Finally we can check for Slant asymptotes. S.A.s exist where the degree of the numerator is 1 greater than the degree of the denominator. In this case the numerator has a degree smaller than the denominator. This means we will not have a slant asymptote.

We look at the zeros of the numerator however we already cancelled out the zero `( x + 3 )` so there will not be any x intercepts

To find the y intercept plug in 0 for x in the remaining equation which is `1 / ( x + 4 )`. The y int will be `1/4`.

SUMMARY

Hole at `( -3, 1 )`

V.A, at `x = -4`

H.A. at `y = 0`

S.A: does not exist

X int: does not exist

Y int: `1/4`