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# Graphing exponential functions | Lesson

## What are exponential functions, and how frequently do they appear on the test?

Note: On your official SAT, you might not see a question about graphing exponential functions at all! At most, you'll see 1 question.
If you haven’t already mastered more frequently tested SAT skills, you may want to save this topic for later.

In an exponential function, the output of the function is based on an expression in which the input is in the exponent. For example, $f\left(x\right)={2}^{x}+1$ is an exponential function, because $x$ is an exponent of the base $2$.
The graphs of exponential functions are nonlinear—because their slopes are always changing, they look like curves, not straight lines:
You can learn anything. Let's do this!

## How do I graph exponential functions, and what are their features?

### Graphing exponential growth & decay

Graphing exponential growth & decaySee video transcript

### Using points to sketch an exponential graph

The best way to graph exponential functions is to find a few points on the graph and to sketch the graph based on these points.
To find a point on the graph, select an input value and calculate the output value. For example, for the function $f\left(x\right)={2}^{x}+1$, if we want to find the $y$-value when $x=1$, we can evaluate $f\left(1\right)$:
$\begin{array}{rl}f\left(1\right)& ={2}^{1}+1\\ \\ & =3\end{array}$
Since $f\left(1\right)=3$, the point $\left(1,3\right)$ is a point on the graph.
We need to use the points to help us identify three important features of the graph:
• What is the $y$-intercept?
• Is the slope of the graph positive or negative?
• What happens to the value of $y$ as the value of $x$ becomes very large?

#### The $y$‍ -intercept

Not only is the $y$-intercept the easiest feature to identify, it also helps you figure out the rest of the features.
To find the $y$-intercept, evaluate the function at $x=0$.
For example, the $y$-intercept of the graph of $f\left(x\right)={2}^{x}+1$ is:
$\begin{array}{rl}f\left(0\right)& ={2}^{0}+1\\ \\ & =2\end{array}$

#### The slope

An exponential function is either always increasing or always decreasing. If you have already evaluated $f\left(0\right)$, try evaluating $f\left(1\right)$.
• If $f\left(1\right)>f\left(0\right)$, then the slope of the graph is positive.
• If $f\left(1\right), then the slope of the graph is negative.
For $f\left(x\right)={2}^{x}+1$, since $f\left(0\right)=2$ and $f\left(1\right)=3$, we can conclude that the slope of the graph is positive because $3>2$.

#### The end behavior

End behavior is just another term for what happens to the value of $y$ as $x$ becomes very large in both the positive and negative directions. For the graph of an exponential function, the value of $y$ will always grow to positive or negative infinity on one end and approach, but not reach, a horizontal line on the other. The horizontal line that the graph approaches but never reaches is called the horizontal asymptote.
For $f\left(x\right)={2}^{x}+1$:
• As $x$ increases, $f\left(x\right)$ becomes very large. The value of $y$ on the right end of the graph approaches infinity.
• As $x$ decreases, $f\left(x\right)$ becomes closer and closer to $1$, but it's always slightly larger than $1$. The value of $y$ on the left end of the graph approaches, but never reaches, $1$.

#### Putting it all together

With the help of a few more points, $\left(-2,1.25\right)$, $\left(-1,1.5\right)$, and $\left(2,5\right)$, we can sketch the graph of $f\left(x\right)={2}^{x}+1$.
Note: if you're graphing by hand, it's more important to recognize that the value of $y$ will grow to positive infinity as $x$ increases than getting the graph exactly right! You can use the points you identified to establish a trend and sketch out the curve.
To graph an exponential function:
1. Evaluate the function at various values of $x$—start with $-1$, $0$, and $1$. Find additional points on the graph if necessary.
2. Use the points from Step 1 to sketch a curve, establishing the $y$-intercept and the direction of the slope.
3. Extend the curve on both ends. One end will approach a horizontal asymptote, and the other will approach positive or negative infinity along the $y$-axis.

Example: Graph $f\left(x\right)=12\cdot {\left(\frac{1}{2}\right)}^{x}$.

### Try it!

try: find points on an exponential graph
$f\left(x\right)={2}^{x}-1$
Identify points on the graph of the exponential function above and completing the table below.
$x$$f\left(x\right)$
$-3$$-0.875$
$-1$
$0$
$1$$1$
$2$

try: describe an exponential graph
$f\left(x\right)={2}^{x}-1$
Using the points from the previous question, complete the following statements about the graph of the exponential function above.
The $y$-intercept of the graph is located at
.
As $x$ increases, $y$
.
As the value of $x$ decreases, the value of $y$ approaches, but never reaches,
.

## How do I identify features of exponential graphs from exponential functions?

### Graphs of exponential growth

Graphs of exponential growthSee video transcript

### Identifying features of graphs from functions

#### The basic exponential function

The most basic exponential function has a base and an exponent:
$f\left(x\right)={b}^{x}$
Let's consider the case where $b$ is a positive real number:
• If $b>1$, then the slope of the graph is positive, and the graph shows exponential growth. As $x$ increases, the value of $y$ approaches infinity. As $x$ decreases, the value of $y$ approaches $0$.
• If $0, then the slope of the graph is negative, and the graph shows exponential decay. In this case, as $x$ increases, the value of $y$ approaches $0$. As $x$ decreases, the value of $y$ approaches infinity.
• For all values of $b$, the $y$-intercept is $1$.
The graphs of $f\left(x\right)={1.5}^{x}$ and $f\left(x\right)={\left(\frac{2}{3}\right)}^{x}$ are shown below.

#### How do we shift the horizontal asymptote?

The $y$-value of every exponential graph approaches positive or negative infinity on one end and a constant on the other. We can change the constant value $y$ approaches by introducing a constant term to the function:
• For $f\left(x\right)={b}^{x}$, the value of $y$ approaches infinity on one end and the constant $0$ on the other.
• For $f\left(x\right)={b}^{x}+d$, the value of $y$ approaches infinity on one end and $d$ on the other.
The graphs of $f\left(x\right)={1.5}^{x}$ and $f\left(x\right)={1.5}^{x}+2$ are shown below.
• For $f\left(x\right)={1.5}^{x}$, as $x$ decreases, the value of $y$ approaches $0$.
• For $f\left(x\right)={1.5}^{x}+2$, as $x$ decreases, the value of $y$ approaches $2$.

#### How do we shift the $y$‍ -intercept?

We can change the $y$-intercept of the graph either by introducing a constant term (as above) or introducing a coefficient for the exponential term:
• For $f\left(x\right)={b}^{x}+d$, the $y$-intercept is $1+d$.
• For $f\left(x\right)=a\cdot {b}^{x}$, the $y$-intercept is $a\cdot 1=a$. In this form, $a$ is also called the initial value.
• For $f\left(x\right)=a\cdot {b}^{x}+d$, the $y$-intercept is $a+d$.
The graphs of $f\left(x\right)={1.5}^{x}+2$, $f\left(x\right)=2\cdot {1.5}^{x}$, and $f\left(x\right)=2\cdot {1.5}^{x}+2$ are shown below.
• For $f\left(x\right)={1.5}^{x}+2$, the $y$-intercept is $1+2=3$.
• For $f\left(x\right)=2\cdot {1.5}^{x}$, the $y$-intercept is $2\cdot 1=2$.
• For $f\left(x\right)=2\cdot {1.5}^{x}+2$, the $y$-intercept is $2+2=4$.

### Try it!

TRY: identify the features of an exponential graph without finding points
$f\left(x\right)=64\left(0.25{\right)}^{x}$
Consider the exponential function $f$ above. The $y$-intercept of its graph, or the initial value of the function, is
.
Because the base of the exponent, $0.25$, is less than $1$, the slope of the graph is
. As the value of $x$ increases by $1$, the value of $y$
.

## You turn!

Practice: match an exponential function to its graph
Which of the following is the graph of the function $y=4\cdot {0.5}^{x}$ ?

Practice: transform an exponential function
The graph of function $f$ is shown in the $xy$-plane above. Which of the following is the graph of $f\left(x\right)+2$ ?

## Things to remember

For $f\left(x\right)={b}^{x}$, where $b$ is a positive real number:
• If $b>1$, then the slope of the graph is positive, and the graph shows exponential growth. As $x$ increases, the value of $y$ approaches infinity. As $x$ decreases, the value of $y$ approaches $0$.
• If $0, then the slope of the graph is negative, and the graph shows exponential decay. In this case, as $x$ increases, the value of $y$ approaches $0$. As $x$ decreases, the value of $y$ approaches infinity.
• For all values of $b$, the $y$-intercept is $1$.
To shift the horizontal asymptote:
• For $f\left(x\right)={b}^{x}$, the value of $y$ approaches infinity on one end and the constant $0$ on the other.
• For $f\left(x\right)={b}^{x}+d$, the value of $y$ approaches infinity on one end and $d$ on the other.
To shift the $y$-intercept:
• For $f\left(x\right)={b}^{x}+d$, the $y$-intercept is $1+d$.
• For $f\left(x\right)=a\cdot {b}^{x}$, the $y$-intercept is $a\cdot 1=a$. In this form, $a$ is also called the initial value.
• For $f\left(x\right)=a\cdot {b}^{x}+d$, the $y$-intercept is $a+d$.

## Want to join the conversation?

• How do you answer the question with only the base and x??
• how do you get the equation of an exponential graph
• how would you graph a number if the x exponet is a diffrent number like negative 3 like for ex: f(X)= 2(3)^x-3 +2 ??
• Hi Angelina,
If I understand your equation correctly, it is 6 to the power of (x-3) plus 2. In other words;
f(x) = 6^(x-3) + 2.
To graph this you would do the same process as the other equations. Plug in a x value, and solve for y.
e.g. x = 4. y = 6^(4 - 3) + 2
y = 6^1 + 2
y = 6 + 2
y = 8

Lastly, if the x value is less than three, then you'll have a negative exponent. This may cause some confusion but don't be afraid as it's easier than it may seem. When a number is to the power of a negative number, it is simply 1 / x^n. Heres an example:
2^(-4)
= 1 / (2^4)
= 1 / 16
I believe there are other khan academy lessons which show this concept.
So using this, we can solve your equation when x is less than 3.
e.g. x = 1
y = 6^(1-3) + 2
y = 6^(-2) + 2
y = (1 / 6^2) + 2
y = (1 / 36) + 2
y = ((1 + 72) / 36)
y = 73 / 36

I hope this helps! :)