Exponential Graphs

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From the course by University of California, Irvine

Pre-Calculus: Functions

291 ratings

University of California, Irvine

Pre-Calculus: Functions

291 ratings

This course covers mathematical topics in college algebra, with an emphasis on functions. The course is designed to help prepare students to enroll for a first semester course in single variable calculus. Upon completing this course, you will be able to: 1. Solve linear and quadratic equations 2. Solve some classes of rational and radical equations 3. Graph polynomial, rational, piece-wise, exponential and logarithmic functions 4. Find integer roots of polynomial equations 5. Solve exponential and logarithm equations 6. Understand the inverse relations between exponential and logarithm equations 7. Compute values of exponential and logarithm expressions using basic properties

From the lesson

Exponential and Logarithmic Equations.

Exponential and logarithmic functions are two important functions that are useful in applications of calculus. We will discuss properties and graphs of these special functions. We will also learn to solve equations involving exponential and logarithmic functions.

Evaluating Logarithmic Expressions 5:31

Solving Logarithmic Equations 6:02

Exponential Graphs 7:29

Logarithmic Graphs 6:55

Meet the Instructors

Dr. Sarah Eichhorn
Associate Dean, Distance Learning & Tenured Lecturer
Dr. Rachel Cohen Lehman

[MUSIC] Let's look at exponential graphs. For example, let's sketch the graph of

f(x) = 3 raised to the x minus 1st power minus 2.

And then we're going to find any x or y-intercepts of its graph.

Let's use graph transformations to help us here.

Let's start by sketching y = 3^x. So, if this is the y-axis and this is the

x-axis, what does y = 3^x look like? Well, it has

a y-intercept at 1 and then, when x is 1, y is 3^1 which is 3.

So, this is 2, here's 3. So, we have the point 1,3 lies on the

graph. And the exponential function looks like

this. Now, the x-axis or y = 0 is a horizontal

asymptote. Now, what does this -1 do here? What that

does is it shifts this graph rigidly 1 unit to the right.

That is the graph of y = 3^x-1 looks like this.

So, here's the y-axis, here's the x-axis. What's going to happen to this point over

here (0,1) if shift our graph one unit to the right?

This is going to move to 0 + (1,1) or (1,1).

And what's going to happen to this point here, 1,3? It's going to move to 1 +

(1,3) or (2 3). Now, let's plot these points.

This is x = 1 and y = 1. Here's our point 1,1 and if this is x =

2, y = 3, then here's our point (2,3).

And the horizontal asymptote will still remain at y = 0, so our graph will look

like this. Now, what does this -2 do to this graph?

What that does is it shifts this entire graph rigidly down two units.

So, let's say this is the y-axis and this is the x-axis.

What is going to happen to this point here, 1,1? It's going to move to 1 and

then 1 - 2 or 1, -1. And what's going to happen to this point

here? This 2,3? It's going to move to (2 3), - 2 or

(2,1). So, let's plot this points over here,

here is x = 1, y = -1.

So, here's our point (1,-1) and then here's 2 and 1.

So, here's our point (2,1). Now the horizontal asymptote was at y =

0, which is also shifting down two units. Which means that the new horizontal

asymptote then, will be at y = -2. And so, our graph looks like this.

And let's darken these points we plotted, that is this is f(x) = 3 raised to x

minus 1st power and then, minus 2. Now, we are also asked to find in the x

or y-intercepts of its graph. And looking at the graph, we see that we

have a y-intercept here and an x-intercept here.

So, let's find these intercepts. To find the y-intercept, we set x = 0 in

our equation. That is, we have y = 3 raised to the 0 -

1 - 2, or y = 3 to the negative first power minus 2, or y = 1/3 - 2, or y =

-5/3, which would be our y-intercept. That is, coming back over here to our

graph, the y-coordinate of this point is -5/3.

Alright. And what about our x-intercept?

To find our x-intercept, we set y = 0. That is, we have zero is equal to 3

raised to the x minus first power minus two or 3 raised to the x minus 1st power

is equal to 2. And now, writing this in the equivalent

logarithmic form gives us that log base 3 of 2 is equal to x - 1.

And now, adding 1 to both sides gives us that x = log base 3 of 2 + 1, which would

be our x-intercept. That is, the x-coordinate of this point

here is log base 3 of 2 + 1. And this is how we work with exponential

graphs. Thank you and we'll see you next time.

[MUSIC]

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