Rational Functions

does the following function have a hole or a vertical asymptote or both? how do you know? find the y-value at that point.

y=x^2+7x+12/x^2+8x+15

I factored this out to

y = (x+3)(x+4)/(x+5)(x+3)
The x+3's cancelled out. I don't get the whole verticle and horizontal asymptote stuff, how do u know w/o using technology whether there is an asymptote and if there is a hole?

If a pair of factors cancel, then you would get a hole

in your case, there is a hole when x = -3
If you have a factor in the denominator which does not cancel, it will cause a vertical asymptote
in your case there will be a vertical asymptote at x = -5

BTW, you should use brackets when typing a function like yours
y=(x^2+7x+12)/(x^2+8x+15)
= (x+3)(x+4)/( (x+5)(x+3) )

there is a vertical asymptote if the denominator is zero and the numerator is not zero

there is a horizontal asymptote if the degree of the numerator is less than or equal to the degree of the denominator

There is a slant asymptote if the degree of the numerator is 1 more than the degree of the denominator.

I can't believe these facts were not covered in your book.

So,

y = (x+4)/(x+5) as long as x ≠ -3

If x = -3, y = 0/0 which is not defined. There is a "hole" at (-3,1/2)

So, you should be able to figure the asymptotes present.

Visit wolframalpha.com and type in

plot (x^2+7x+12)/(x^2+8x+15)

you can play around with lots of stuff there

Revived after 9 years xD

To determine if a rational function has a hole or a vertical asymptote, you need to look for factors that cancel out in both the numerator and the denominator. In this case, you correctly factored the function as:

y = (x+3)(x+4)/(x+5)(x+3)

As you noticed, the (x+3) terms cancel out, which means there is a hole at x = -3 in the graph of the function. To find the y-value at that point, you can substitute x = -3 into the simplified function.

When x = -3:

y = (-3+3)(-3+4)/(-3+5)(-3+3)
= 0/2
= 0

So, the y-value at the hole (x = -3) is 0.

Now, to determine if there is a vertical asymptote, you need to check if there are any factors in the denominator that do not cancel out with any factors in the numerator. In this case, the factor (x+5) in the denominator does not cancel out with any factors in the numerator. Therefore, there is a vertical asymptote at x = -5.

In summary, the function has a hole at x = -3 and a vertical asymptote at x = -5. The y-value at the hole is 0.