# Exponential Functions

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Lesson Priority: VIP Knowledge

Objectives
• Apply recently acquired knowledge about exponential relationships to the input/output behavior we are familiar with for functions
• Know and understand the form of a basic exponential function, $f(x) = ab^x$
• Describe common properties of exponential functions, and continue to understand the difference between growth and decay
• Interpret the coefficients of an exponential function, and what happens when the coefficient is positive vs negative
• Using growth or decay and a positive or negative coefficient, learn and understand the four major categories that an exponential function can fall into: positive growth, positive decay, negative growth, and negative decay
• See the general shape of exponential graphs, and how that shape changes in each of the four exponential categories
Lesson Description

This lesson will put together several ideas that we saw in the prior three lessons, defining exponential relationships in the context of looking at a function. Not only will be employ techniques for solving for a variable like we saw in the prior lesson, but we will also define general properties of an exponential function based on the coefficients in the typical exponential form we see.

Practice Problems

Practice problems and worksheet coming soon!

## Applying the Function Concept

Functions are all about looking at the relationship between input values and corresponding output values. This is no different for functions with variable exponents, but the properties of this family of functions are fundamentally different from polynomial, radical, rational, or trigonometric functions that you may have studied up to this point.Defining Exponential FunctionsAs we recently explored in our introductory lesson on exponential relationships », relationships with variable exponents fall into two categories decided by their base number $b$. If $b$ is bigger than $1$, the relationship is called growth, while relationships with fraction values of $b$ are decay. We'll build on that further here as we go from relationships to functions, and see what other common features these types of functions have.First, let's define exponential functions including function notation.
Define: Exponential FunctionsFunctions are considered exponential if the input variable (usually $x$) is an exponent.$$f(x) = ab^x + k$$e.g.$$g(x) = 5^x$$or$$h(x) = 2\cdot 3^{x} - 1$$
Note that $b$ must be positive for us to think of it as a "regular" function. Technically, if $b$ was negative, it would still be a function by the book definition of what a function is, but exponentials with negative bases aren't studied because they are not "compact". Compactness is something you would study in advanced real function analysis courses, so feel free to look it up if you're curious, but the takeaway is that you will not be studying functions with negative base values.$b$ must also not be $1$, since $1$ to any power is itself, which is a constant number function.

## Doing Function Things

Plugging InAs we often do with functions, we'll need to be able to evaluate at specified $x$ values.

Example 1For the exponential function$$f(x) = 2\cdot 3^x - 1$$Determine $f(0)$, $f(4)$, and $f(-2)$.$\blacktriangleright$ Plug in these values for $x$ to get each answer:\begin{align} f(0) & = 2\cdot 3^0 - 1 \\ & = 2\cdot 1 - 1 = 1 \end{align}\begin{align} f(4) & = 2 \cdot 3^4 - 1 \\ & = 2 \cdot 81 - 1 = 161 \end{align}\begin{align} f(-2) & = 2 \cdot 3^{-2} - 1 \\ & = 2 \cdot \frac{1}{9} - 1 = -\frac{7}{9} \end{align}

You Should Know
We're very often allowed to use a calculator to evaluate specific values of exponential functions. However, when working with integers, you may be expected to mentally compute results, including negative exponents (e.g. $3^{-2} = 1/9$).
We could also be asked to figure out what value of $x$ causes the function to be equal to a certain result.

Example 2Determine the value of $x$ that makes $g(x)=28$, where$$g(x) = 5^{-x} + 3$$$\blacktriangleright$ Like we've seen in the past when doing this task, we need to set the function equal to $28$ and solve for $x$$5^{-x} + 3 = 28$$$$5^{-x} = 25$$From here, we know that$5$must be raised to the power$2$, and so$$-x = 2$$$$\therefore x = -2$$ You Should Know Solving for$x$when a function is set equal to a particular value often requires logarithms » but we can answer these questions without them if the solution is an integer or simple fraction. GraphingWhile the next lesson is dedicated to creating exponential graphs », we need to first become familiar with what they look like.Recently we discovered that variable exponent relationships » come in four flavors: positive growth, negative growth, positive decay, and negative decay. It is important that you know which is which visually, and why.Positive growthNegative growthPositive decayNegative decayWe'll also want to understand how these graphs change when we apply function transformations, which we'll address further down the page.End BehaviorUnderstanding end behavior goes hand in hand with graph familiarity. What's important to understand is that in each of the four types of graphs, the side that quickly turns upward or downward approaches infinity or negative infinity as you examine$x$values farther and farther left or right. Also, the opposite side levels off and gets closer and closer to zero (in the absence of transformations). ## Determining an Exponential Function Occasionally, it will be our job to figure out what the exponential function is, given that we know which points it passes through.Without a vertical shift transformation, an exponential function will have the form$$f(x) = ab^x$$and since there are two numbers involved, there are two unknowns to solve for. Example 3Determine the exponential function$f(x) =ab^x$that passes through the points$\displaystyle\left(2,\frac{25}{2}\right)$and$\displaystyle\left(-1,\frac{27}{10}\right)$.$\blacktriangleright$Given these points, we know$$f(2) = \frac{25}{2}$$and$$f(-1) = \frac{27}{10}$$Because we know the function is of the form$ab^x$, we can substitute that knowledge in both cases:$$ab^2 = \frac{25}{2}$$$$ab^{-1} = \frac{27}{10}$$In order to figure out what$a$and$b$each are, we can look at these two equations as a system. However, unlike linear systems » of equations that you solved in Algebra, the way to solve this kind of system is with division.Let's divide the equations by one another. This works as an extension of the transitive property of equality, because if an equation means that the things on each side have the same value, then dividing the left-hand side pieces by one another should be equal in value to what you get when you divide the right-hand side pieces by one another.$$ab^2 = \frac{25}{2}$$$$ab^{-1} = \frac{27}{10}$$$$\longrightarrow$$$$\frac{ab^2}{ab^{-1}} = \frac{\frac{25}{2}}{\frac{27}{10}}$$Cleaning up by cancelling the$a$terms on the left and re-writing the division of fractions more simply on the right,$$\frac{b^2}{b^{-1}} = \frac{25}{2} \cdot \frac{10}{27}$$$$\frac{b^2}{b^{-1}} = \frac{125}{27}$$Now, we can simplify the left side by using exponent rules for dividing terms of the same base:$$b^{2-(-1)} = \frac{125}{27}$$$$b^3 = \frac{125}{27}$$Taking the cube root of each side yields us with b:$$b = \frac{5}{3}$$To get$a$, pick either equation and plug in$b$. For example, we can use the first of the two:$$ab^2 = \frac{25}{2}$$$$\longrightarrow a\left(\frac{5}{3}\right)^2 = \frac{25}{2}$$$$a \cdot \left( \frac{25}{9} \right) = \frac{25}{2}$$Multiply both sides by$9/25$to finish the problem:$$a = \frac{9}{2}$$Therefore, the function we were looking for is$$f(x) = \frac{9}{2} \cdot \left( \frac{5}{3} \right)^x$$ ## Common Function Properties Although exponential functions may be transformed via vertical shifts (more on that in a few paragraphs), we are more often than not examining functions of the form$f(x) = ab^x$. Here are some quick good-to-know properties about this family of functions.The Base is PositiveWe will never spend any meaningful time on functions with a negative$b$value, because that type of relationship lacks several important properties that most functions have. Even if$b$is a fraction, it will always be positive.We won't bother working with the function if$b$is$1$because$1$to any power remains one, and so$f(x) = a(1)^x$becomes$f(x) = a \cdot 1$which is no longer an exponential relationship.b Determines Growth or DecayWhen$b$is a fraction between$0$and$1$, the exponential relationship will be classified as decay. This means that as$x$becomes larger, the function value will shrink in magnitude (overall height of graph), ultimately converging toward zero.When$b$is any value greater than$1$, the relationship will be classified as exponential growth. This means that as$x$becomes larger, the function will quickly increase in magnitude, and do so at an increasing rate. Remember! We saw in the last lesson » that teachers can try and use negative exponents to trick us. If$b$is raised to the negative$x$, this is equivalent to using the reciprocal of$b$with a positive$x$. In other words:$$b^{-x} = \left(\frac{1}{b}\right)^x$$This is a very common way to confuse students between growth and decay.$y=(1/2)^x$is a decay function but$y=(1/2)^{-x}$is a growth function. Similarly,$y = 5^x$is a growth function but$y=5^{-x}$is a decay function. a Determines Vertical OrientationWe have learned about growth and decay in a few instances, and when pairing those categories with whether$a$value is positive or negative, we have the complete picture of the four outcomes we described above: positive growth, negative growth, positive decay, and negative decay. For$f(x) = ab^x$, the word positive means that the function will be entirely above the$x$axis, while negative means that the function will be entirely below the$x$axis.Once again, here are the four general graph shapes.Positive growthNegative growthPositive decayNegative decayThe$a$value is the only trigger for whether or not the function is negative because$b^x$on its own is always positive, since$b>0$, even if$b$is a fraction. ## Function Transformations Regardless of whether an exponential function is positive or negative, growth or decay, we know that all functions follow the same transformation rules » when it comes to seeing how parent functions change graphically in response to particular algebraic changes.Thanks to the particular structure of exponential functions, the transformation that we need to be most aware of is vertical translations (sometimes referred to as shifts or movements). The other common types of transformations (translations in the horizontal direction as well as both vertical and horizontal stretch or compressions) do not change the general shape and pattern of an exponential relationship. We'll explore why in the final paragraphs of this lesson (Mr. Math Makes It Mean »).Vertical shifts change the asymptote in the direction that does not grow infinitely largely. The parent function$f(x) = ab^x$has a horizontal asymptote of$y=0$in that direction, but the graph of$g(x) = ab^x + k$has an asymptote of$y=k$.For the other shifts and compression / stretch behaviors, everything follows along with what we already know, but the graph will look very similar to the parent function.For a parent function$f(x) = ab^x$, the following transformations are associated with these corresponding algebraic changes.Horizontal Shift:$f(x) = ab^{x-h}$Vertical Compression or Stretch:$f(x) = c\cdot ab^x$Horizontal Compression or Stretch:$f(x) = ab^{cx}$The fact that these transformations do not substantially change the look of the graph can actually be a challenge, because teachers often expect you to label enough points on the two so you can tell them apart. We'll practice drawing exponential graphs » from scratch in the next lesson, which is when it will be important to remember this.For now, examine the following transformations of the parent function$f(x) = 2^x$and make sure you can see why each graph is the way it is.Vertical Shift:Horizontal Shift:Vertical Stretch:Horizontal Compression: ## Mr. Math Makes It Mean Equating TransformationsUnique to exponential functions, applying a horizontal shift transformation gives a result that could be achieved identically by instead applying a vertical stretch or compression. One of the reasons this is challenging is that teachers expect you to show and communicate the "concept" difference even though they are the same function. For example, here are the graphs of$g(x) = (0.25)2^x$and$h(x) = 2^{x-2}$, each graphed against their parent function$f(x) = 2^x$:The other reason this comes up is that teachers can actually ask you to find and describe the matching pairs of transformations. Example 4For the parent function$h(x) = 6^x$, transform it into a new function$f(x)$by shifting it left$3$units. Then find a vertical scale transformation of$h(x)$to create a new function$g(x)$that achieves the same result.$\blacktriangleright$Let's start by applying what we know about horizontal transformations. Moving this function$3$units to the left gives us the new function that we'll call$f(x)$:$$f(x) = 6^{x+3}$$This satisfies the first question. Next, let's find$g(x)$which is a pure vertical scale transformation of$h(x)$, but is algebraically equivalent to$f(x)$. Do this by starting with$f(x)$and using exponent rules.$$f(x) = 6^{x+3}$$The exponent rule$x^{a} \cdot x^{b} = x^{a+b}$, used in reverse, tells us that$$\longrightarrow f(x) = 6^x \cdot 6^3 = 216 \cdot 6^x$$Therefore, we can say that$g(x) = 216 \cdot 6^x$is equal to$f(x) = 6^{x+3}$, and so we have found two functions$f$and$g$, each of which is the desired transformation of the parent function, that are equal to one another algebraically. Reflections and Growth / DecayWhen working with advanced transformations like reflections, and when multiple transformations are in play, it can be difficult to understand the properties of an exponential function. We'll continue to practice multiple transformations in the next lesson on graphing, but right away we should understand the order of operations when multiple changes are made to an exponential function. Example 5Take a look at the following function$g(x)$derived from the parent function$f(x) = 2^x$:$$g(x) = -3\cdot 2^{5-x} - 4$$Identify what transformations are happening, and what properties of the function we know.$\blacktriangleright$There's a lot going on here, but identifying what transformations occur is not enough - the order matters. Work inside-out, thinking of the first things you would do to$x$if you had a value for it that you were trying to plug in.First we'd deal with the$5-x$piece. This is one of the trickiest parts of the problem, because$5-x$can be written as$-(x-5)$, meaning we're really dealing with a negative exponent - this is actually a decay function! In function transformation speak, this means we have a horizontal shift to the right of$5$, and then a horizontal reflection, when comparing to the parent function. With graphs in the next lesson, it will be more clear that a horizontal reflection will turn a growth parent function into a decay function, and vice versa.Finishing the problem - staying in order of arithmetic operations for plugging in a value of$x$, we know that the next thing after calculating$2^{-(x-5)}$would be applying the$-3$multiplier, which is a vertical reflection and a vertical stretch of$3$. Finally, the$-4$is a vertical shift of$4$downward.Answering the question concisely:To obtain$g(x)$from$f(x)$, apply the following transformations to$f(x)$, in this order: • Horizontal shift right$5$units • Horizontal reflection • Vertical stretch by a factor of$3$• Vertical reflection • Vertical shift downward$4$units Additionally,$g(x)$is a negative decay function. So what's the takeaway of working with multiple transformations? 1. Be able to identify not only the transformations applied, but also the order in which they apply. 2. Know that any reversed exponent like$5-x$really needs to be written and looked at as$-(x-5)$, to understand the nature of which horizontal direction shift happens before the horizontal refection occurs. ## Put It To The Test Example 6Fill in the following$x$-$y$chart for the given function$f(x)$.$$f(x) = \frac{1}{2} \cdot 4^x$$$$\begin{array}{|c|c|} \hline x & y \\ \hline -2 & \\ \hline -1 & \\ \hline 0 & \\ \hline 2 & \\ \hline 5 & \\ \hline \end{array}$$ Show solution$\blacktriangleright$Find$f(x)$for each$x$value in the table.$$f(-2) = \frac{1}{2} \cdot 4^{-2} = \frac{1}{2} \cdot \frac{1}{16} = \frac{1}{32}$$$$f(-1) = \frac{1}{2} \cdot 4^{-1} = \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8}$$$$f(0) = \frac{1}{2} \cdot 4^{0} = \frac{1}{2} \cdot 1 = \frac{1}{2}$$$$f(2) = \frac{1}{2} \cdot 4^{2} = \frac{1}{2} \cdot 16 = 8$$$$f(5) = \frac{1}{2} \cdot 4^{5} = \frac{1}{2} \cdot 1024 = 512$$The complete table looks like:$$\begin{array}{|c|c|} \hline x & y \\ \hline -2 & \frac{1}{32} \\ \hline -1 & \frac{1}{8} \\ \hline 0 & \frac{1}{2} \\ \hline 2 & 8 \\ \hline 5 & 512 \\ \hline \end{array}$$ Example 7Solve for the value of$x$that makes$f(x) = 54$for the function$$f(x) = \frac{3}{2} \cdot 6^x$$ Show solution$\blacktriangleright$Let's set up the information in an equation. We don't know$x$but we know when we plug in$x$to the function, we get$54$.$$\longrightarrow \frac{3}{2} \cdot 6^x = 54$$Multiplying both sides by$\frac{2}{3}$yields$$6^x = 36$$$$\longrightarrow x = 2$$Note that you may have been able to solve by guess and check since the answer was an integer. It's actually difficult for teachers to create non-integer non-calculator questions for exponential functions. Example 8Determine whether the following exponential functions are growth or decay, positive or negative.$$f(x) = 2 \cdot \left(\frac{5}{3}\right)^x$$$$g(x) = \frac{1}{2} \cdot \left(\frac{3}{8}\right)^{-x}$$$$h(x) = -4^x$$$$j(x) = 10 \cdot \left(\frac{1}{10}\right)^{x}$$$$k(x) = \frac{5}{2} \cdot 7^x$$ Show solution$\blacktriangleright$Exponential functions are called negative or positive only based on their leading multiplier (if one exists). Growth and decay are labels determined by the base number$b$, but remember that negative exponents flip the base and therefore flip the label.With that in mind:$f(x)$is positive growth$g(x)$is positive growth$h(x)$is negative growth$j(x)$is positive decay$k(x)$is positive growth Example 9Write an exponential function in the form$g(x) = ab^x$such that$g(-3) = 1280$and$g(1) = 5$Show solution$\blacktriangleright$Start by plugging in the$x$values into the function definition:$$g(x) = ab^x$$$$g(-3) \longrightarrow ab^{-3} = 1280$$$$g(1) \longrightarrow ab^{1} = 5$$Take the two equations on the right and divide one by the other. The laws of exponents help us simplify the left-hand sides.$$\frac{ab^{-3}}{ab^{1}} = \frac{1280}{5}$$The$a$'s cancel, leaving$$b^{-4} = 256$$Raise both sides to the power$-1/4$to neutralize the exponent:$$b = \frac{1}{4}$$Finally, to get$a$, use either equation and plug in$b$. The second equation is a little simpler to use.$$a \cdot \left(\frac{1}{4}\right)^1 = 5$$$$\longrightarrow a = 20$$Therefore, the unknown function is$$g(x) = 20 \cdot \left(\frac{1}{4}\right)^x$$$$g(x) = 20\cdot \left(\frac{1}{4}\right)^x$$ Example 10For the parent function$$f(x) = \left(\frac{1}{3}\right)^x$$Determine the transformations that need to be applied to$f(x)$in order to obtain the new function$g(x)$, where$$g(x) = \frac{1}{2} \cdot \left(\frac{1}{3}\right)^{3-x} + 2$$ Show solution$\blacktriangleright$As we've seen with multiple transformations, we need to use PEMDAS to guide our thinking. Pretend you are plugging in a specific value for$x$, and consider what arithmetic you would evaluate in what order.The other minor trick here is that we really need to re-write the$3-x$exponent as$-(x-3)$, so that we can properly label the translation and the reflection.With that in mind, the transformations we need to apply to$f(x)$to obtain$g(x)$are, in order: • Horizontal shift to the right$3$units • Horizontal reflection • Vertical scale (compression) by$1/2$• Vertical shift up$2\$ units

Lesson Takeaways
• Know the usual form of exponential functions
• Continue to gain familiarity with the growth, decay, positive, and negative qualities that these functions can have
• Know the common properties of exponential functions relative to polynomial or rational functions you've seen in the past
• Work with transformations of exponential functions
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