Solving and Graphing Exponential Equations - Algebra II

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Question

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

Answer

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.

$-2e^{x-3} < -2\cdot 0$

$-2e^{x-3} < 0$

$-2e^{x-3} -5 < 0-5$

$-2e^{x-3} -5 < -5$

So the value that cannot exceed is , and the line is the asymptote.

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Question

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

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

Answer

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 .

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Question

Find the vertical asymptote of the equation.

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

Answer

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

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

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Question

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

Answer

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.

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Question

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

Answer

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:

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Question

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

Answer

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 .

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Question

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

Answer

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:

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Question

Determine whether each function represents exponential decay or growth.

$ewline a) ewline y=(\frac{1}{2})^{x}$

Answer

a)

This is exponential decay since the base, , is between and .

b)

This is exponential growth since the base, , is greater than .

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Question

Give the -intercept of the graph of the equation $f(x) = 2 ^{x- 7} + 4$ .

Answer

Set and solve for

$f(x) = 2 ^{x- 7} + 4$

$2 ^{x- 7} + 4 = 0$

$2 ^{x- 7} = -4$

We need not work further. It is impossible to raise a positive number 2 to any real power to obtain a negative number. Therefore, the equation has no solution, and the graph of has no -intercept.

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Question

What is/are the asymptote(s) of the graph of the function $g(x) = 0.4 ^{x - 2} + 0.7$ ?

Answer

An exponential function of the form

$g(x) = a^{x- h}+k$

has as its one and only asymptote the horizontal line .

Since we define as

$g(x) = 0.4 ^{x - 2} + 0.7$ ,

then ,

and the only asymptote is the line of the equation .

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Question

Match each function with its graph.

1.

2.

3.

a.

b.

c.

Answer

For , our base is greater than so we have exponential growth, meaning the function is increasing. Also, when , we know that since . The only graph that fits these conditions is .

For , we have exponential growth again but when , . This is shown on graph .

For , we have exponential decay so the graph must be decreasing. Also, when , . This is shown on graph .

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Question

In 2010, the population of trout in a lake was 416. It has increased to 521 in 2015.

Write an exponential function of the form that could be used to model the fish population of the lake. Write the function in terms of , the number of years since 2010.

Answer

We need to determine the constants and . Since in 2010 (when ), then and

To get , we find that when , . Then $b^5=\frac{521}{416}$ .

Using a calculator, ${\left(\frac{521}{416}\right)}^{\frac{1}{5}}=1.046$ , so $b\approx 1.046$ .

Then our model equation for the fish population is

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Question

An exponential funtion is graphed on the figure below to model some data that shows exponential decay. At , is at half of its initial value (value when ). Find the exponential equation of the form $y=Ae^{kt}$ that fits the data in the graph, i.e. find the constants and .

Answer

To determine the constant , we look at the graph to find the initial value of , (when ) and find it to be . We can then plug this into our equation $y=Ae^{kt}$ and we get $1.2=Ae^{k0}$ . Since $e^{k0}=e^0=1$ , we find that .

To find , we use the fact that when , is one half of the initial value . Plugging this into our equation with now known gives us $\frac{1.2}{2}=0.6=1.2e^{k(0.3648)}$ . To solve for , we make use the fact that the natural log is the inverse function of , so that

$\ln(e^{k(0.3648)})=k(0.3648)$ .

We can write our equation as $\frac{0.6}{1.2}=e^{k(0.3648)}$ and take the natural log of both sides to get:

$\ln(\frac{0.6}{1.2})=\ln(e^{k(0.3648)})$ or .

Then $k\approx -1.9$ .

Our model equation is $y=1.20e^{-1.9t}$ .

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Question

What is the -intercept of the graph $y = 3^{x}$ ?

Answer

The -intercept of any graph describes the -value of the point on the graph with a -value of .

Thus, to find the -intercept substitute .

In this case, you will get,

$y = 3^{0} = 1$ .

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Question

What is the -intercept of $y = 4(2^{x})$ ?

Answer

The -intercept of a graph is the point on the graph where the -value is .

Thus, to find the -intercept, substitute and solve for .

Thus, we get:

$y = 4(2^{0}) = 4(1)=4$

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Question

What is the -intercept of $y = 5^{x+2}$ ?

Answer

The -intercept of any function describes the point where .

Substituting this in to our funciton, we get:

$y = 5^{x+2} = 5^{0+2} = 5^{2} = 25$

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Question

Which of the following functions represents exponential decay?

Answer

Exponential decay describes a function that decreases by a factor every time increases by .

These can be recognizable by those functions with a base which is between and .

The general equation for exponential decay is,

where the base is represented by and .

Thus, we are looking for a fractional base.

The only function that has a fractional base is,

$y = \left(\frac{1}{3}\right)^{x}$ .

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Question

Does the function $y = 4^{x}$ have any -intercepts?

Answer

The -intercept of a function is where . Thus, we are looking for the -value which makes $0 = 4^{x}$ .

If we try to solve this equation for we get an error.

To bring the exponent down we will need to take the natural log of both sides.

Since the natural log of zero does not exist, there is no exponent which makes this equation true.

Thus, there is no -intercept for this function.

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Question

Which of the following correctly describes the graph of an exponential function with a base of three?

Answer

Exponential functions with a base greater than one are models of exponential growth. Thus, we know that our function will increase and not decrease. Remembering the graph of an exponential function, we can determine that the graph will begin gradually, almost like a flat line. Then, as increases, begins to increase very quickly.

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Question

Solve the equation for .

$\small 9^x=3^6$

Answer

Begin by recognizing that both sides of the equation have a root term of .

$\small 9^x=3^6$

Using the power rule, we can set the exponents equal to each other.

$3^{(2*x)}=3^6$

$\small 2x=6$

$\small x=3$

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