Algebra II - Asymptotes

What is/are the asymptote(s) of the graph of the function

$f(x) = 5e^{x-2}-2$ ?

Explanation

An exponential equation of the form $f(x) = ae^{x-h} +k$ has only one asymptote - a horizontal one at . In the given function, , so its one and only asymptote is .

Consider the exponential function . Determine if there are any asymptotes and where they lie on the graph.

There is one horizontal asymptote at .

There are no asymptotes. goes to positive infinity in both the and directions.

There is one vertical asymptote at .

Explanation

For positive values, increases exponentially in the direction and goes to positive infinity, so there is no asymptote on the positive -axis. For negative values, as decreases, the term becomes closer and closer to zero so approaches as we move along the negative axis. As the graph below shows, this is forms a horizontal asymptote.

Exp_asymp

Determine the asymptotes, if any: $y=\frac{x^2-2x+1}{x^2-1}$

$x=\pm1$

$\textup{There are no asymptotes.}$

$x=-\frac{1}{2}$

Explanation

Factorize both the numerator and denominator.

$y=\frac{x^2-2x+1}{x^2-1}=\frac{(x-1)(x-1)}{(x+1)(x-1)}$

Notice that one of the binomials will cancel.

The domain of this equation cannot include $x=\pm1$ .

The simplified equation is:

$y=\frac{x-1}{x+1}$

Since the term canceled, the term will have a hole instead of an asymptote.

Set the denominator equal to zero.

Subtract one from both sides.

There will be an asymptote at only:

The answer is:

Where is an asymptote located, if any? $y=\frac{2x-4}{x^2-4}$

$\textup{There is no asymptote.}$

Explanation

Factor the numerator and denominator.

Rewrite the equation.

$y=\frac{2x-4}{x^2-4}=\frac{ 2(x-2)}{(x+2)(x-2)}=\frac{2}{x+2}$

Notice that the will cancel. This means that the root of will be a hole instead of an asymptote.

Set the denominator equal to zero and solve for x.

An asymptote is located at:

The answer is:

Find the vertical asymptote of the equation.

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

There are no vertical asymptotes.

Explanation

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

$\sqrt{16}=\sqrt{x^2}$

Which of the choices represents asymptote(s), if any? $y=\frac{2x^2+2x}{x^2-8x-9}$

$x=1, 9\textup{ and } y= -\frac{2}{5}$

$\textup{There are no asymptotes.}$

$x=9\textup{ and }y=\frac{2}{5}$

Explanation

Factor the numerator and denominator.

$y=\frac{2x^2+2x}{x^2-8x-9} = \frac{2x(x+1)}{(x-9)(x+1)}$

Notice that the terms will cancel. The hole will be located at because this is a removable discontinuity.

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

The denominator cannot be equal to zero. Set the denominator to find the location where the x-variable cannot exist.

The asymptote is located at .

What is the horizontal asymptote of the graph of the equation $y = -2e^{x-3} -5$ ?

$y=-\frac{5}{2}$

Explanation

The asymptote of this equation can be found by observing that $e^{x-3} > 0$ regardless of . We are thus solving for the value of as $e^{x-3}$ approaches zero.