# Find the horizontal and vertical asymptotes of the curve.

• Jan 31st 2008, 05:52 PM
plstevens
Find the horizontal and vertical asymptotes of the curve.
• Jan 31st 2008, 06:45 PM
The vertical asymptotes will appear when the denominator is 0. So you need to solve $\displaystyle x^2+x-2 = 0$

The horizontal asymptote is the value for y when x approaches $\displaystyle \infty$ or $\displaystyle -\infty$. Can you do it now?
• Jan 31st 2008, 08:15 PM
mr fantastic
Quote:

[snip]
The horizontal asymptote is the value for y when x approaches $\displaystyle \infty$ or $\displaystyle -\infty$.

In other words, you consider the two limits:

$\displaystyle y = \lim_{x \rightarrow +\infty} \frac{7x^2 + x - 9}{x^2 + x - 2}$, and

$\displaystyle y = \lim_{x \rightarrow -\infty} \frac{7x^2 + x - 9}{x^2 + x - 2}.$

In this case, the limits have the same value.
• Feb 4th 2008, 09:22 PM
plstevens
how do u find this value
• Feb 4th 2008, 09:26 PM
mr fantastic
Quote:

Originally Posted by mr fantastic
In other words, you consider the two limits:

$\displaystyle y = \lim_{x \rightarrow +\infty} \frac{7x^2 + x - 9}{x^2 + x - 2}$, and

$\displaystyle y = \lim_{x \rightarrow -\infty} \frac{7x^2 + x - 9}{x^2 + x - 2}.$

In this case, the limits have the same value.

$\displaystyle y = \lim_{x \rightarrow +\infty} \frac{7x^2 + x - 9}{x^2 + x - 2} = \lim_{x \rightarrow +\infty} \frac{7 + \frac{1}{x} - \frac{9}{x^2}}{1 + \frac{1}{x} - \frac{2}{x^2}}= \frac{7 + 0 - 0}{1 + 0 - 0} = .....$.