Math - Understanding Asymptotes

What are the -intercepts of the equation?

$y=\frac{x^2-9}{x^2-16}$

There are no -intercepts.

Explanation

To find the x-intercepts of the equation, we set the numerator equal to zero.

$\sqrt{9}=\sqrt{x^2}$

$y=\frac{x^2-9}{x^2-16}$

What is the horizontal asymptote of this equation?

There is no horizontal asymptote.

Explanation

Look at the exponents of the variables. Both our numerator and denominator are . Therefore the horizontal asymptote is calculated by dividing the coefficient of the numerator by the coefficient of the denomenator.

$y=\frac{1}{1}=1$

$y=\frac{3x^2-5}{2x^2+7}$

What are the horizontal asymptotes of this equation?

$y=\frac{3}{2}$

There are no horizontal asymptotes.

$y=\frac{2}{3}$

Explanation

Since the exponents of the variables in both the numerator and denominator are equal, the horizontal asymptote will be the coefficient of the numerator's variable divided by the coefficient of the denominator's variable.

For this problem, since we have $\frac{3x^2}{2x^2}$ , our asymptote will be $y=\frac{3}{2}$ .

$y=\frac{3x^2-5}{2x^2+7}$

What are the vertical asymptotes of the equation?

There are no real vertical asymptotes.

There are no vertical asymptotes.

$x=-\sqrt{\frac{7}{2}}$

$x=-\sqrt{\frac{7}{2}}, \sqrt{\frac{7}{2}}$

$x=\frac{7}{2}$

Explanation

To find the vertical asymptotes, we set the denominator equal to zero and solve.

$0={2x^2+7}$

$\frac{-7}{2}=x^2$

$\sqrt{-\frac{7}{2}}=\sqrt{x^2}$

$\sqrt{-\frac{7}{2}}=x$

Since we'd be trying to find a negative number, we have no real solution. Therefore, there are no real vertical asymptotes.

Find the vertical asymptote(s) of $y=\frac{x^3+2x+8}{x^2-36}$ .

and

There are no real vertical asymptotes.

$x=\frac{3}{2}$

Explanation

To find the vertical asymptotes, we set the denominator of the fraction equal to zero, as dividing anything by zero is "undefined." Since it's undefined, there's no way for us to graph that point!

Take our given equation, $y=\frac{x^3+2x+8}{x^2-36}$ , and now set the denominator equal to zero:

$\sqrt{36}=\sqrt{x^2}$

Don't forget, the root of a positive number can be both positive or negative ( as does ), so our answer will be $\pm6=x$ .

Therefore the vertical asymptotes are at and .

$y=\frac{x-3}{x^2-12}$

What are the vertical asymptotes of the equation?

$x=2\sqrt{3},-2\sqrt{3}$

$x=\sqrt{12}$

There are no vertical asymptotes.

Explanation

To find the vertical asymptotes, we set the denominator equal to zero.

$\sqrt{12}=\sqrt{x^2}$

$\sqrt{4*3}=x$

Because the square root only gives us the absolute value, our answer will be:

$x=2\sqrt{3}, -2\sqrt{3}$

Find the vertical asymptote(s) of $y=\frac{x^2+9x+8}{x^2-44}$ .

$x=2\sqrt{11}$ and $x=-2\sqrt{11}$

$x=2\sqrt{11}$

and

There are no real vertical asymptotes for this function.

Explanation

To find the vertical asymptotes, we set the denominator of the fraction equal to zero, as dividing anything by zero is undefined.

Take our given equation, $y=\frac{x^2+9x+8}{x^2-44}$ , and now set the denominator equal to zero:

$\sqrt{44}=\sqrt{x^2}$

is not a perfect square, but let's see if we can pull anything out.

$\sqrt{4*11}=\sqrt{x^2}$

$\sqrt{4}*\sqrt{11}=\sqrt{x^2}$

$2\sqrt{11}=x$

Don't forget that there is a negative result as well:

$x=-2\sqrt{11}$ .

Find the horizontal asymptote(s) of $y=\frac{x^3+2x+8}{x^2-36}$ .

There are no real horizontal asymptotes.

$y=\frac{3}{2}$

and

Explanation

To find the horizontal asymptote of the function, look at the variable with the highest exponent. In the case of our equation, $y=\frac{x^3+2x+8}{x^2-36}$ , the highest exponent is in the numerator.

When the variable with the highest exponent is in the numberator, there are NO horizontal asymptotes. Horizontal asymptotes only appear when the greatest exponent is in the denominator OR when the exponents have same power in both the denominator and numerator.

$y=\frac{x^2-9}{x^2-16}$