Topics
6. Exponential & Logarithmic Functions
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Algebra 2

6.01 Exponential functions

Lesson
Practice

Characteristics of exponential functions

An exponential relationship can be modeled by a function with the independent variable in the exponent, known as an exponential function:

\displaystyle f\left(x\right)=ab^x
\bm{a}
Leading coefficient
\bm{b}
Base where b>0, b\neq 1
\bm{x}
Independent variable
\bm{f\left(x\right)}
Dependent variable

We can also call b the growth or decay factor.

Exploration

Move the sliders for a and b to see how the base and the exponent affect the graph of the exponential function.

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  1. What happens to the graph as b increases?
  2. What happens to the graph as b is between 0 and 1?
  3. What happens to the graph as a decreases?
  4. Describe the end behavior for:
    • a>0 and b>1?
    • a>0 and 0<b<1?
    • a<0 and b>1?
    • a<0 and 0<b<1?
  5. Create two equations where the function is increasing over its domain.
  6. Create two equations where the function is decreasing over its domain.

The general shape for different values of a and b is shown.

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y>0, Increases at an increasing rate
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All exponential functions of the form y=ab^x have the following features in common:

  • The domain is -\infty <x<\infty.
  • The y-intercept is at \left(0,a\right).
  • There is a horizontal asymptote at y=0.
Horizontal asymptote

A line that a curve or graph approaches as it heads toward positive or negative infinity.

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Examples

Example 1

Which graph(s) show an exponential function? Select all correct graphs.

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Worked Solution
Create a strategy

Look at the shape of the graph. An exponential graph will have a horizontal asymptote which one end will be approaching, while the other end tends to infinity.

Apply the idea

Graph B is a straight line, indicating a linear graph.

Graph C has a U-shape, representing a quadratic graph.

Graphs A and D are exponential curves, with A showing a>0 and 0<b<1 and D showing a<0 and b>1.

So, the correct answers are options A and D.

Example 2

Consider the two functions, f\left(x\right) = 2 \left(\dfrac{1}{3}\right)^x and g\left(x\right) = x^2+2.

a

Complete the table of values and sketch f\left(x\right) = 2 \left(\dfrac{1}{3}\right)^x.

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Worked Solution
Create a strategy

To find the table of values for the function f\left(x\right) = 2 \left(\dfrac{1}{3}\right)^x, substitute different values of x into the equation and calculate the corresponding y values. Then, plot these points on a graph to sketch the function.

Apply the idea

Evaluate f(-2) : f(-2)=2 \left(\dfrac{1}{3}\right)^{-2}=2 \cdot (3)^{2}=18

Evaluate f(-1) : f(-1)=2 \left(\dfrac{1}{3}\right)^{-1}=2 \cdot (3)^{1}=6

Evaluate f(0) : f(0)=2 \left(\dfrac{1}{3}\right)^{0}=2 \cdot 1=2

Evaluate f(1) : f(1)=2 \left(\dfrac{1}{3}\right)^{1}=2 \cdot \dfrac{1}{3}=\dfrac{2}{3}

Evaluate f(2) : f(2)=2 \left(\dfrac{1}{3}\right)^{2}=2 \cdot \dfrac{1}{9}=\dfrac{2}{9}

Table of values:

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y1862\dfrac{2}{3}\dfrac{2}{9}

Graph:

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b

Sketch the graph of g\left(x\right) on the same plane as f\left(x\right).

Worked Solution
Create a strategy

To sketch the graph of g\left(x\right) = x^2+2, we can use the transformation of the parent function x^2, which is a parabola. The graph of x^2 is a parabola with its vertex at the origin and opening upward.

Apply the idea

Since g\left(x\right) = x^2 + 2, the transformation from the parent function x^2 is a vertical shift upward by 2 units. This means that the vertex of the parabola for g\left(x\right) is at \left(0, 2\right). The shape of the parabola remains the same as the parent function, opening upward.

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c

Compare the y-intercepts of both functions.

Worked Solution
Create a strategy

The y-intercept of a function is the point where it intersects the y-axis. We can find the y-intercepts from the graphs and compare their values.

Apply the idea
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From the graphs, we can see that the y-intercept of f\left(x\right) is at the point (0,2) and the y-intercept of g\left(x\right) is at the point (0,2) as well.

Both functions have the same y-intercept.

Reflect and check

Another way to find the y-intercepts of the functions is to substitute x=0 directly into their equations and evaluate them.

\displaystyle f(0)\displaystyle =\displaystyle 2 \left(\dfrac{1}{3}\right)^0Substitute x=0 into f\left(x\right)
\displaystyle f(0)\displaystyle =\displaystyle 2(1)Evaluate the exponent
\displaystyle f(0)\displaystyle =\displaystyle 2Evaluate the multiplication
\displaystyle g(0)\displaystyle =\displaystyle (0)^2+2Substitute x=0 into g\left(x\right)
\displaystyle g(0)\displaystyle =\displaystyle 2Evaluate

This confirms that both functions have the same y-intercept.

d

Compare the end behavior of both functions.

Worked Solution
Create a strategy

To compare the end behavior, we need to look at what happens to the y-values of each function as x approaches positive and negative infinity.

Apply the idea
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For f\left(x\right), as x gets larger and larger, the function gets closer and closer to the asymptote causing the y-values of the function to approach 0. \text{As }x\to\infty,f\left(x\right)\to 0

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As x gets smaller and smaller, the y-values of the function get very large and approach positive infinity.\text{As } x\to-\infty, f\left(x\right)\to\infty

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For g\left(x\right), as x decreases and increases, the y-values get very large, causing both ends of the graph to approach positive infinity.\begin{aligned}&\text{As }x\to-\infty, g\left(x\right)\to\infty\\&\text{As }x\to\infty, g\left(x\right)\to\infty\end{aligned}

We can describe the end behavior of the functions in mathematical notation as:

As x \to -\infty , {f\left(x\right) \to \infty} and as x \to \infty , {f\left(x\right) \to 0}

As x \to -\infty , {g\left(x\right) \to \infty} and as x \to \infty , {g\left(x\right) \to \infty}

We can see that both functions have the same behavior on the left as x approaches negative infinity, but their behavior on the right as x approaches positive infinity differs.

Reflect and check

To reasonably verify the behavior of the functions f\left(x\right) and g\left(x\right) algebraically, you can use a calculator to input very large and very small values for x.

For f\left(x\right), we can substitute a small value , like x=-100, into f\left(x\right) which gives us f\left(-100\right) = 1.0308 \cdot 10^{48}. This is a very large number which matches our end behavior of:\text{As }x \to -\infty,f\left(x\right) \to \infty

We can also substitute a large value, like x=100, into f\left(x\right) which gives us f\left(100\right) = 3.8807 \cdot 10^{-48}. This is a number very close to zero which matches our end behavior of:\text{As }x \to \infty,f\left(x\right) \to 0

For g\left(x\right), we can substitute a small value, like x=-100, into g\left(x\right) which gives us g\left(-100\right) = 10002. This is a very large number which matches our end behavior of:\text{As }x \to -\infty,g\left(x\right) \to \infty

We can also substitute a large value, like x=100, into g\left(x\right) which gives us g\left(100\right) = 10002. This is a very large number which matches our end behavior of:\text{As }x \to \infty,g\left(x\right) \to \infty

e

Compare the domain and range of both functions.

Worked Solution
Create a strategy

We can use the graphs to find the domain and range of each function and compare them. The domain of a function is the set of all possible x-values, and the range is the set of all possible y-values.

Apply the idea
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The graph of the exponential function f\left(x\right) , never touches the x-axis, but it exists for every value of x, so its domain is all real numbers, or (-\infty, \infty).

The graph has an asymptote at y=0, but the graph exists for every y-value above the asymptote. This gives us a range of (0, \infty).

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The quadratic function g\left(x\right), also exists for all x-values, so its domain is all real numbers, or (-\infty, \infty).

It has a vertex at the point (0,2), which is the lowest point of the graph. g\left(x\right) exists at its vertex and for all y-values above the vertex so the range is all real numbers greater than or equal to 2, or [2, \infty).

Comparing the domain and range of both functions, we can see that both functions have the same domain, (-\infty, \infty), but their ranges are different.

Idea summary

Exponential functions can take the form:

\displaystyle f\left(x\right)=ab^x
\bm{a}
Leading coefficient
\bm{b}
Base where b>0, b\neq 1
\bm{x}
Independent variable
\bm{f\left(x\right)}
Dependent variable

Exponential functions of the form y=ab^x have the following features in common:

  • The domain is -\infty <x<\infty.
  • The y-intercept is at (0,a).
  • There is a horizontal asymptote at y=0.

Transformations of exponential functions

To draw the graph of an exponential function, we can use a variety of strategies, including:

  • Completing a table of values for the function and drawing the curve through the points found
  • Using transformations
  • Identifying key features from the equation
  • Using technology, such as a physical or online graphing calculator

Exploration

Move the sliders to see how each one affects the graph.

Loading interactive...
  1. What happens to the graph when k is positive or negative?
  2. What happens to the graph when h is positive or negative?
  3. What happens to the graph when a is positive or negative?
  4. What happens to the graph when c is positive or negative?
  5. What impact does c have on the graph?
  6. What happens to the graph when 0<b<1?

For the function:

\displaystyle f\left(x\right)=a\cdot b^{c\left(x-h\right)}+k
\bm{b}
is the base
\bm{a}
can reflect, stretch or compress vertically
\bm{c}
can reflect, stretch or compress horizontally
\bm{h}
determines the horizontal translation
\bm{k}
determines the vertical translation
\text{Reflection across the }x\text{-axis:}y=-b^x
\text{Reflection across the }y\text{-axis:}y=b^{-x}
\text{Vertical stretch when} \left|a\right|>1 \\ \text{Vertical compression when } 0<\left|a\right|<1 \text{:}y=a\left(b\right)^x
\text{Horizontal compression when} \left|c\right|>1 \\ \text{Horizontal stretch when} 0<\left|c\right|<1 \text{:}y=b^{c\cdot x}
\text{Horizontal translation by } h \\ \text{Vertical translation by } k \text{:}y=b^{x-h}+k

For horizontal stretches and compressions, recall c=\dfrac{1}{\text{scale factor}}

The parent function y=b^x has key features:

  • Increasing for b>1, and decreasing for 0<b<1
  • Domain of \left(-\infty, \infty\right) and range of \left[0, \infty\right)
  • y=0 is the horizontal asymptote
  • y-intercept at \left(0,1\right)
  • Another point at \left(1,b\right)

To get a more precise shape, we can map the asymptote and the points \left(0,1\right) and \left(1,b\right) using the transformations.

Examples

Example 3

For each graph, first identify the transformation from the parent function f\left(x\right)=\left(\dfrac{1}{3}\right)^x, then write the equation of g\left(x\right).

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Worked Solution
Create a strategy

Compare the graph of the parent function with the graph of the transformed function. Notice that the parent function is shifted to the right to create g\left(x\right).

Apply the idea
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This shift can be identified by selecting a point on the function f\left(x\right) and observing the movement of its corresponding point on g\left(x\right).

For example, the point f\left(0\right)=1 corresponds to g\left(3\right)=1 on the new function.

This shows that the x-value moved 3 units to the right while the y-value remained unchanged, indicating a horizontal shift of 3 units in the positive direction.

A horizontal translation is represented by g\left(x\right)=\left(\dfrac{1}{3}\right)^{x-h}. We have just found that h=3. Therefore, the equation of the transformed function isg\left(x\right)=\left(\dfrac{1}{3}\right)^{x-3}

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Worked Solution
Create a strategy

Compare f\left(x\right) and g\left(x\right). Notice that it appears they are the same size and shape, but g\left(x\right) is a mirror image of f\left(x\right) over the x-axis.

Apply the idea
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Let's identify points on f\left(x\right) and compare them to the corresponding points on g\left(x\right).

For example, f\left(-1\right)=3 corresponds to g\left(-1\right)=-3.

For each point on f\left(x\right), the corresponding point on g\left(x\right) shares the same x-value, but the y-value is negative. This indicates a reflection across the x-axis.

In the function notation g\left(x\right)=a\cdot f\left(x\right), a reflection across the x-axis is represented by a=-1.

Therefore, the equation of g\left(x\right) is g\left(x\right) = -\left(\dfrac{1}{3}\right)^x

Reflect and check

To check our solution, we can find points on f\left(x\right) and g\left(x\right) algebraically and compare them.

Let's compare the points on f\left(x\right) and g\left(x\right) when x=1.

When we substitute x=1 into f\left(x\right), we get f\left(1\right) = \left( \dfrac{1}{3}\right)^1= \dfrac{1}{3}. This gives us the point \left(1,\dfrac{1}{3}\right).

When we substitute x=1 into g\left(x\right), we get g\left(1\right) = - \left(\dfrac{1}{3}\right)^1=-\dfrac{1}{3}. This gives us the point \left(1,-\dfrac{1}{3}\right).

When we compare these two points, we see that they share the same x-value, but the y-value of g\left(x\right) is the negative of f\left(x\right). We can do this with every possible x-value and get the same result where the y-values of g\left(x\right) are the negative of the y-values of f\left(x\right).

This confirms that we have a reflection about the x-axis.

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Worked Solution
Create a strategy

Compare f\left(x\right) and g\left(x\right). Notice that they have the same horizontal asymptote and y-intercept, so a translation has not occurred. The shape of the functions are different, indicating a dilation has occurred.

Apply the idea
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Similar to the previous examples, let's consider the corresponding points for f\left(-1\right) and f\left(-2\right).

f\left(-1\right)=3 corresponds to g\left(-2\right)=3. Also, f(-2)=9 corresponds to g\left(-4\right)=9.

For each point on f\left(x\right), the corresponding point on g\left(x\right) has the same y-value, but the x-value has been doubled. This indicates a horizontal stretch by a factor of 2.

A horizontal dilation is represented by g\left(x\right)=\left(\dfrac{1}{3}\right)^{c\cdot x} where c=\dfrac{1}{\text{scale factor}}. We found the scale factor to be 2, so c=\dfrac{1}{2}.

Therefore, the equation of g\left(x\right) is g\left(x\right)=\left(\dfrac{1}{3}\right)^{0.5x}

Reflect and check

It can sometimes be difficult to identify whether the dilation was vertical or horizontal, but checking more than one point on the graph can help us avoid mistakes.

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For example, we may have mistakenly said the corresponding point of f\left(-2\right)=9 was g\left(-2\right)=3. This would correspond to a vertical shrink by a factor of \dfrac{1}{3}.

We can now check whetherf\left(-1\right)=3 was also compressed by \dfrac{1}{3}. However, g\left(-1\right)\approx \dfrac{4}{3} which corresponds to a scale factor of \dfrac{4}{9}.

Since these scale factors are not the same, we know that the graph was not dilated vertically.

Example 4

Given the parent function f\left(x\right) = 4^x, graph g\left(x\right).

a

g\left(x\right) = 4^{-x}

Worked Solution
Create a strategy

We can use a table of values to graph the parent function, then identify the transformation that occurred to create g\left(x\right). Then, we can use the transformation to graph g\left(x\right).

Apply the idea

First, create a table of values to graph the parent function f\left(x\right) = 4^x:

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f\left(x\right)\dfrac{1}{16}\dfrac{1}{4}1416
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Notice that the only change between the parent function, f\left(x\right) = 4^x, and the transformed function, g\left(x\right) = 4^{-x}, is the exponent of g\left(x\right) is negative. This indicates a reflection across the y-axis.

By reflecting each point of f\left(x\right) across the y-axis, we can create the graph of g\left(x\right):

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Reflect and check

Using technology, we can check that we have correctly identified and graphed the transformation.

A screenshot of the GeoGebra Graphing tool showing the graphs of f of x equals 4 raised to x and g of x equals 4 raised to negative x. Talk to your teacher for more information.

Notice that the graph of g\left(x\right) is a reflection of f\left(x\right) across the y-axis. We can also see that the points on our graph match the points shown in the graphing calculator. This verifies our solution.

b

g\left(x\right) = -4^{x}+3

Worked Solution
Create a strategy

The parent function is f\left(x\right) = 4^x, and the transformed function is g\left(x\right) = -4^{x}+3. In the transformation notation, y=a\left(b\right)^{x-h}+k, the negative sign corresponds to a=-1 and constant of 3 corresponds to k=3.

Recall that the a-value indicates a vertical reflection or dilation, and the k-value indicates a vertical translation.

Apply the idea
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When a=-1, a reflection across the x-axis has occurred. This means that the graph of g\left(x\right) will be a mirrored version of the graph of f\left(x\right) with respect to the x-axis.

Begin by graphing f\left(x\right), then reflect each point to the other side of the x-axis.

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Since k=3, the original function has been translated up 3 units.

To graph this tranformation, we will move each point on the reflected graph up 3 units.

Note that translating the function upwards will also translate the horizontal asymptote. The asymptote was y=0, so it will now be y=3.

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Reflect and check

Recall that when graphing transformations, we always graph reflections and dilations before translations. If we had graphed the translation first, we would have gotten a different graph.

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This shows f\left(x\right) after it has been translated up 3 units.

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Now, applying the reflection to the translated function, we would get the graph shown.

Notice that this graph is not the same as the one we found, which shows that the order in which we apply the transformations matters.

c

g\left(x\right) = - \left(\dfrac{1}{4}\right)\cdot 4^x

Worked Solution
Create a strategy

The parent function is f\left(x\right) = 4^x, and the transformed function is g\left(x\right) = - \left(\dfrac{1}{4}\right)\cdot 4^x. The differences between the two functions are the negative sign and the factor \dfrac{1}{4} in front of the exponential term. Both of these correspond to the a-value in the transformation notation y=a\left(b\right)^{x-h}+k.

Apply the idea

For g\left(x\right) = - \left(\dfrac{1}{4}\right)\cdot 4^x, a=-\dfrac{1}{4}. Since a<0, the function has been reflected across the x-axis. Also, 0<\left|a\right|<1, so the function has been compressed vertically by a factor of \dfrac{1}{4}.

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f\left(x\right)\dfrac{1}{16}\dfrac{1}{4}1416
g\left(x\right)\dfrac{1}{64}\dfrac{1}{16}\dfrac{1}{4}14

To apply the vertical dilation, we will multiply each point on f\left(x\right) by \dfrac{1}{4}.

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Plotting these points helps use see that the dilation of the graph of f\left(x\right) by a factor of \dfrac{1}{4}.

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Next, we will reflect each point across the x-axis. This means that the dilated graph will be a mirrored version of the graph with respect to the x-axis.

Note that the horizontal asymptote will remain at y=0.

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Reflect and check

The order in which we apply reflections and dilations does not matter. If we had reflected the function first, then dilated it, we would have gotten the same graph. Again, we can verify our solution using technology.

A screenshot of the GeoGebra Graphing tool showing the graphs of f of x equals 4 raised to x and g of x equals negative one fourth times 4 raised to x. Talk to your teacher for more information.

Example 5

Consider the following exponential functions:

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g\left(x\right)-9-3-1-\dfrac{1}{3}-\dfrac{1}{9}
a

Determine which function increases at a slower rate.

Worked Solution
Create a strategy

To help us visualize the functions, we can graph them on the same coordinate plane.

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Apply the idea

When we look at the graph, we can see g\left(x\right) does not seem as steep as f\left(x\right) and its curve is more rounded. This is a visual indication that g\left(x\right) increases at a slower rate.

Reflect and check

Since the functions have not been translated, another way to determine the answer is by comparing the constant factors of the functions. We can find the constant factors by dividing the outputs of two consecutive values of x.

For f\left(x\right), we can use the points (-1, -8) and (0,-2).

b=\dfrac{-2}{-8}=\dfrac{1}{4}

For g\left(x\right), we can use the points (-2, -9) and (-1, -3).

b=\dfrac{-3}{-9}=\dfrac{1}{3}

Since \dfrac{1}{4}<\dfrac{1}{3}, f\left(x\right) gets multiplied by a smaller factor than g\left(x\right). Because of this, f\left(x\right) will get closer to 0 faster, so g\left(x\right) increases at a slower rate.

b

Identify the y-intercept for each function.

Worked Solution
Create a strategy

On the given graph of f\left(x\right), we will look for the point where the function intercepts the y-axis. In the table for g\left(x\right), we will look for the point where x=0.

Apply the idea

The y-intercept of f\left(x\right) is (0,-2).

The y-intercept of g\left(x\right) is (0,-1).

Reflect and check

Now that we know the y-intercept and constant factor for each function, we can build the equations.

f\left(x\right)=-2\left(\dfrac{1}{4}\right)^x

g\left(x\right)=-1\left(\dfrac{1}{3}\right)^x

c

Identify and compare f(-1) and g(-1).

Worked Solution
Create a strategy

To find f(-1), we can use the graph to locate the point (-1,y). For g(-1), we can use the table of values to find the corresponding y-value when x = -1. Then, we can compare the two functions.

Apply the idea
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Looking at the graph, we will find the point on the function when x=-1. This occurs at (-1, -8), therefore, f(-1)=-8.

Using the table of values for the function g\left(x\right), we can find the value of g(-1) by locating the row with x = -1 and reading its corresponding y-value. When x = -1,\, g\left(x\right) = -3.

We find that f(-1)=-8 and g(-1)=-3 which are different values and g(-1)> f(-1).

Idea summary

We can use transformations to graph an exponential function in the form {y=a \cdot b^{c (x-h)} + k} and identify key features:

\text{Reflection across the }x\text{-axis:}y=-b^x
\text{Reflection across the }y\text{-axis:}y=b^{-x}
\text{Vertical stretch when} \left|a\right|>1 \\ \text{Vertical compression when } 0<\left|a\right|<1 \text{:}y=a\left(b\right)^x
\text{Horizontal compression when} \left|c\right|>1 \\ \text{Horizontal stretch when} 0<\left|c\right|<1 \text{:}y=b^{c\cdot x}
\text{Horizontal translation by } h \\ \text{Vertical translation by } k \text{:}y=b^{x-h}+k

For horizontal stretches and compressions, recall c=\dfrac{1}{\text{scale factor}}

The parent function y=b^x has key features:

  • Increasing for b>1, and decreasing for 0<b<1
  • Domain of \left(-\infty, \infty\right) and range of \left[0, \infty\right)
  • y=0 is the horizontal asymptote
  • y-intercept at \left(0,1\right)
  • Another point at \left(1,b\right)

Inverses of exponential functions

Recall that when inverse functions are graphed on the same Cartesian plane, they are reflections of each other about the line y=x.

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Each point \left(x,y\right) on the graph of f\left(x\right) corresponds to the point \left(y,x\right) on f^{-1}\left(x\right). The key features are affected in a similar way.

  • The y-intercept of f\left(x\right) becomes the x-intercept of f^{-1}\left(x\right).

  • The x-intercept of f\left(x\right) becomes the y-intercept of f^{-1}\left(x\right).

  • The horizontal asymptote of f\left(x\right), y=a, becomes the vertical asymptote of f^{-1}\left(x\right), x=a.

  • The domain of f\left(x\right) becomes the range of f^{-1}\left(x\right).

  • The range of f\left(x\right) becomes the domain of f^{-1}\left(x\right).

Examples

Example 6

Consider the graph of 3^x.

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a

Sketch the inverse function on the same coordinate plane.

Worked Solution
Create a strategy

Swap the x and y-values on the curve of y=3^x to find the corresponding points on the curve of the inverse function.

Apply the idea
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Based on the graph of y=3^x, the points (0,1) and (1,3) lie on the curve of y=3^x.

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The inverse points corresponding to (0,1) and (1,3) are (1,0) and (3,1), respectively. So, these points can be used to sketch the graph of the inverse function of y=3^x.

b

Compare the key features of f\left(x\right) and its inverse including intercepts, zeros, asymptotes, domain, and range.

Worked Solution
Create a strategy

Use the graphs from part (a) to compare the y-intercepts, zeros, asymptotes, domain, range, and any increasing or decreasing intervals. Remember that the inverse function is created by swapping the inputs and outputs, so the features related to inputs and outputs will be affected in a similar way.

Apply the idea
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By observing the graphs of both functions, we can now compare their key features:

  1. y-intercepts: The point where the graph crosses the y-axis.

    • f\left(x\right): y=1
    • g\left(x\right): none
  2. Zeros: Zeros are the x-intercepts of the graph.
    • f\left(x\right): none
    • g\left(x\right): x=1

  3. Asymptotes:

    • f\left(x\right): has a horizontal asymptote at y = 0 but no vertical asymptote.
    • g\left(x\right): has a vertical asymptote at x = 0 but no horizontal asymptote

  4. Domain: The set of all possible x-values for which the function is defined.

    • f\left(x\right): all real numbers, (-\infty, \infty)
    • g\left(x\right): (0, \infty)
  5. Range: The set of all possible y-values for which the function is defined.
    • f\left(x\right): (0, \infty)
    • g\left(x\right): all real numbers, (-\infty, \infty)

  6. Increasing and Decreasing: Both functions are always increasing on their domain. They are never decreasing.

In general, the features of the original function are the opposite of the corresponding features on the inverse function.

Idea summary

Inverse functions are reflections of each other about the line y=x. Each point \left(x,y\right) on the graph of f\left(x\right) corresponds to the point \left(y,x\right) on f^{-1}\left(x\right). The key features are affected in a similar way.

  • The y-intercept of f\left(x\right) becomes the x-intercept of f^{-1}\left(x\right).

  • The x-intercept of f\left(x\right) becomes the y-intercept of f^{-1}\left(x\right).

  • The horizontal asymptote of f\left(x\right), y=a, becomes the vertical asymptote of f^{-1}\left(x\right), x=a.

  • The domain of f\left(x\right) becomes the range of f^{-1}\left(x\right).

  • The range of f\left(x\right) becomes the domain of f^{-1}\left(x\right).

Outcomes

A2.F.1

The student will investigate, analyze, and compare square root, cube root, rational, exponential, and logarithmic function families, algebraically and graphically, using transformations.

A2.F.1a

Distinguish between the graphs of parent functions for square root, cube root, rational, exponential, and logarithmic function families.

A2.F.1b

Write the equation of a square root, cube root, rational, exponential, and logarithmic function, given a graph, using transformations of the parent function, including f(x) + k; f(kx); f(x + k); and kf(x), where k is limited to rational values. Transformations of exponential and logarithmic functions, given a graph, should be limited to a single transformation.

A2.F.1c

Graph a square root, cube root, rational, exponential, and logarithmic function, given the equation, using transformations of the parent function including f(x) + k; f(kx); f(x + k); and kf(x), where k is limited to rational values. Use technology to verify transformations of the functions.

A2.F.1e

Compare and contrast the graphs, tables, and equations of square root, cube root, rational, exponential, and logarithmic functions.

A2.F.2

The student will investigate and analyze characteristics of square root, cube root, rational, polynomial, exponential, logarithmic, and piecewise-defined functions algebraically and graphically.

A2.F.2a

Determine and identify the domain, range, zeros, and intercepts of a function presented algebraically or graphically, including graphs with discontinuities.

A2.F.2b

Compare and contrast the characteristics of square root, cube root, rational, polynomial, exponential, logarithmic, and piecewise-defined functions.

A2.F.2c

Determine the intervals on which the graph of a function is increasing, decreasing, or constant.

A2.F.2f

For any value, x, in the domain of f, determine f(x) using a graph or equation. Explain the meaning of x and f(x) in context, where applicable.

A2.F.2g

Describe the end behavior of a function.

A2.F.2h

Determine the equations of any vertical and horizontal asymptotes of a function using a graph or equation (rational, exponential, and logarithmic).

A2.F.2j

Graph the inverse of a function as a reflection over the line y = x.

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