Question

Find the horizontal asymptote:
7x^2-3x-9/2x^2-4x+5

Answer 1

My guess is 4 and 3? Maybe?

Answer 2

horizontal asymptotes deal with how the function acts for large values of x

Answer 3

for sufficiently large values of x (negative or postive) the function gets controlled by the higher degree terms in such a manner that the other terms become miniscule

Answer 4

\[Lim (x \rightarrow \infty )(\frac{7x^2-3x-9}{2x^2-4x+5})=Lim (x \rightarrow \infty )(\frac{7x^2}{2x^2})=3.5\]Eyeballing it,

Answer 5

Sot it's what comes between the two in a sense?

Answer 6

The vertical asymptotes are the values of x that make the denominator zero but the denominator of this equation doesn't factor and there are no vertical asymptotes. To find the horizontal asymptotes you have to look at the largest exponents in both the numerator and denominator.

If the numerator's largest power is bigger than the denominator's largest power, there is no horizontal asymptote.
If the numerator's largest power is smaller than the denominator's largest, there is a horizontal asymptote at y=0
In this case, the numerator and the denominator have the same power (x^2) so look at the coefficients of the two x^2 terms. The horizontal asymptote is y=7/2 or 3.5.