---
# System prepended metadata

title: 5.2 - Exponential Functions and Graphs
tags: ['5.2']

---

# 5.2 - Exponential Functions and Graphs (J curves)


- Graph of $$f(x)=2^x$$



### Example 1.  Graph $f(x)=2^x$




$$
\begin{array}{|c|c|}
\hline
x & y = f(x) = 2^x \\
\hline
1 & 2^1 = 2 \\
\hline
2 & 2^2 = 4 \\
\hline
3 & 2^3 = 8 \\
\hline
4 & 2^4 = 16 \\
\hline
0 & 2^0 = 1 \\
\hline
-1 & 2^{-1} = \frac{1}{2}=0.5 \\
\hline
-2 & 2^{-2} = \frac{1}{4}=0.25 \\
\hline
-3 & 2^{-3} = \frac{1}{8}=0.125 \\
\hline
-4 & 2^{-4} = \frac{1}{16}=0.0625 \\
\hline
\end{array}
$$
![{47D5106F-481F-462C-BE98-D216A0FC53A1}](https://hackmd.io/_uploads/ByuQZHECC.png)

### Example 2. Graph $f(x)=2^{-x}$

$$
\begin{array}{|c|c|}
\hline
x & y = f(x) = 2^{-x} = \left(\frac{1}{2}\right)^x \\
\hline
1 & 2^{-1} = \frac{1}{2} \\
\hline
2 & 2^{-2} = \frac{1}{4} \\
\hline
3 & 2^{-3} = \frac{1}{8} \\
\hline
4 & 2^{-4} = \frac{1}{16} \\
\hline
0 & 2^0 = 1 \\
\hline
-1 & 2^{1} = 2 \\
\hline
-2 & 2^{2} = 4 \\
\hline
-3 & 2^{3} = 8 \\
\hline
-4 & 2^{4} = 16 \\
\hline
\end{array}
$$

![{F0915020-90D9-44A3-9CD1-A99ADD046A68}](https://hackmd.io/_uploads/HJDdfr4A0.png)

### Example 3. Graph both $f(x)=2^x$ and $g(x)=3^x$

$$
\begin{array}{|c|c|c|}
\hline
x & 2^x & 3^x \\
\hline
1 & 2 & 3 \\
\hline
2 & 4 & 9 \\
\hline
3 & 8 & 27 \\
\hline
4 & 16 & 81 \\
\hline
0 & 1 & 1 \\
\hline
-1 & \frac{1}{2} & \frac{1}{3} \\
\hline
-2 & \frac{1}{4} & \frac{1}{9} \\
\hline
-3 & \frac{1}{8} & \frac{1}{27} \\
\hline
-4 & \frac{1}{16} & \frac{1}{81} \\
\hline
\end{array}
$$

![{110603F2-513C-43CF-91E7-09DE2E843B2B}](https://hackmd.io/_uploads/Sk9iXHVRC.png)

### Graph $y=-2^x$

$$
\begin{array}{|c|c|}
\hline
x & y = f(x) = -2^x \\
\hline
1 & -2^1 = -2 \\
\hline
2 & -2^2 = -4 \\
\hline
3 & -2^3 = -8 \\
\hline
4 & -2^4 = -16 \\
\hline
0 & -2^0 = -1 \\
\hline
-1 & -2^{-1} = -\frac{1}{2} \\
\hline
-2 & -2^{-2} = -\frac{1}{4} \\
\hline
-3 & -2^{-3} = -\frac{1}{8} \\
\hline
-4 & -2^{-4} = -\frac{1}{16} \\
\hline
\end{array}
$$

![{D33DEA31-4010-4D36-9B69-1E14A69D062D}](https://hackmd.io/_uploads/SyuI4rV00.png)


- $f(x)=a^x$ is an **exponential** function.
- The value of $a$ is the **base**, $x$ the exponent.
- The number $e$ is an important base in science and finance:

$$e \approx 2.718281828459 \ldots$$
- Power functions are of the type $x^2, x^3,x^4 \ldots$

- if $x>0$, 
$$2^x<e^x<3^x<4^x<5^x< \ldots$$



## Compound interest.

Let  
- $P$ = the initial amount of money in an account (known as the **principal**).  
- $r$ = the annual **interest rate** as a decimal.  
- $n$ = the number of times the interest is added ("compounded") per year.  
- $t$ = the number of years the account runs.  
- $A$ = the final amount.

Then

$$A = P \left(1 + \frac{r}{n}\right)^{nt}$$


**Example.** We begin with \$800, and compound it quarterly at an annual rate of 8% for 6 years.  
What is the final amount?

- Quarterly means four times per year.
- Convert the interest rate to a decimal by shifting the decimal point 2 places to the left.

We have:

$$P = 800$$

$$r = 0.08$$

$$n = 4$$

$$t = 6$$

So:

$$A = P\left(1 +\dfrac{r}{n}\right)^{nt}$$

$$A = 800\left(1 + \frac{0.08}{4}\right)^{(4)(6)}$$

$$A = 800(1.02)^{24}\approx 1286.75$$


Thus the final amount is $1286.75


---

**Example.** We begin with \$1200, and compound it semi-annually at an annual rate of 6% for $t$ years.  
What is the final amount?

- Semi-annually means two times per year.
- Convert the interest rate to a decimal by shifting the decimal point 2 places to the left.

We have:

$$P = 1200$$

$$r = 0.06$$

$$n = 2$$

$$t = t$$

So:

$$A = P\left(1 +\dfrac{r}{n}\right)^{nt}$$

$$A = 1200\left(1 +\dfrac{0.06}{2}\right)^{2t}$$

$$A=1200(1.03)^{2t}$$

![image](https://hackmd.io/_uploads/S1WChd40R.png)

