Lesson Objectives:- Graph exponential equations and exponential functions.
- Solve applied problems involving exponential functions.
An exponential function with base `a` is `f(x) = a^x`, where x is in the exponent, and it's a real number. And `a > 0`, to avoid imaginary numbers. And `a != 1`, to avoid the identity function; since `1^x` would equal 1.
Sketch the graph of the function, `f(x) = 3^(x+1)`.
So start by setting up a table with your x and y-values. Let's input -2, -1, 0, 1, and 2. When x is -2, `y = 3^(-2+1)`, which is `3^(-1)` or `1/3`. When x is -1, `y = 3^(-1+1)`, which is `3^0`, or just 1. When x is 0, `y = 3^(0+1)`, or `3^1` which is 3. When x is 1, `y = 3^(1+1)`, which is `3^2`, or 9. And when x is 2, `y = 3^(2+1)`, which is equal to `3^3`, or 27.
So our graph of f contains the points: `(-2, 1/3)`, (-1, 1), (0, 3), (1,9), and (2, 27), to name a few. And all that's left to do is plot the points and complete the graph. So our graph looks like this.
The graph of `f(x) = 3^(x+1)` is the graph of `f(x) = 3^x` shifted to the left 1 unit.
When Charles was born, his grandparents gave him a $5,000 certificate of deposit that earns 3.5% interest, compounded quarterly. If the certificate of deposit matures on his 18th birthday, what amount will be available then?
The compound interest formula is `A = P(1+r/n)^(nt)`. A is the final amount, P is the starting amount, r is the interest rate in decimal form, n is the amount of times per year that it's compounded, and t is the number of years.
So we're given 5000 as P, and we're given a rate of 3.5%. If we convert 3.5% to decimal form, we get 0.035. And since it's compounded quarterly, we put 4 for n. And since it matures on his 18th birthday, we know that `t = 18`.
And now we solve for A. So `A = 9362.36`. On his 18th birthday, $9,362.36 will be available.
Sketch the graph of the funciton `f(x) = e^(3x)`.
e was a number which was introduced by a Leonard Euler in 1741. It's equal to 2.71828, etc., and you can find this value on your calculator.
So let's start with a table of our x and our y-values. We'll input -3, -2, -1, 0, and 1. And when we put -3 in, we get 0.00012. And -2 gives us 0.00248. And -1 gives us 0.04979. And 0 gives us 1, and 1 gives us 20.086. So now if we plot these points and complete the graph, we get the following graph.
Notice that the graph of `f(x) = e^(3x)` is the same as `f(x) = e^x`, shrunk horizontally by a factor of 3.