Question

Find the vertical and horizontal asymptote of x^2+3x+3/x^2+5x+4

Answer 1

The vertical asymptotes (and any restrictions on the domain) come from the zeroes of the denominator, so set the denominator equal to zero and solve.

Answer 2

compare the degrees of the numerator and the denominator

1. if the degrees are the same, then you have a horizontal asymptote at y = (numerator's leading coefficient) / (denominator's leading coefficient)
2. if the denominator's degree is greater (by any margin), then you have a horizontal asymptote at y = 0 (the x-axis)
3. if the numerator's degree is greater (by a margin of 1), then you have a slant asymptote which you will find by doing long division

Answer 3

The degrees are the same so the horizontal asymptote is 1?

Answer 4

that is right, good job!

Answer 5

And the vertical is x=-4?

Answer 6

And theres a zero at -2?

Answer 7

there are actually two vertical asymptotes, you are right about x=-4, but there is one more. and yes, there is a zero at x=-2

Answer 8

@jeweaver, you might like to explain what the curve crosses the horizontal asymptote

Answer 9

Is there a vertical asymptote at x=-1 also?

Answer 10

that is correct well done. :)