Common Functions - Online Tutor, Practice Problems & Exam Prep

Graphs of Common Functions

Hey, everyone. So up to this point, we've been spending a lot of time focusing on functions. We've been looking at the graphs of functions and finding the domain and range. And in this video, we're going to be taking a look at some of the common functions. Now, these are the functions that are going to frequently show up throughout this course.

It's definitely important to make sure we're familiar with these functions and their graphs and just in general how they work. So let's take a look at the constant function. The constant function occurs when our function f(x) is equal to c, and c can be any constant number. So, let's say for example our function f(x) is equal to 2. This would be a constant function because 2 is just a constant, and so notice on our graph that y = 2 is consistently the value in all directions. Now notice how this value expands to all the negative and all the positive x's. So we can input any x that we want into this function. We would say that the domain goes from negative infinity to positive infinity because we can have any x value we want. But, notice how the range is only going to be where our y value is 2. So y = 2 is the only possible output we can get for this function. That's the domain and range of the constant function, and keep in mind that this is just going to be whatever c is. So it's not always going to be 2; it could be 4, it could be negative 5/3.

Now let's take a look at the identity function next. The identity function says that f(x) is equal to x, and this function tells you that whatever you put into the function, you're going to get out of it. So let's say that you put negative one in for x. Well, in this case, you're gonna get negative one as your output.

Now, let's say you put 50 in for x. Well, in this case, you're gonna get 50 as your output. So whatever you put in, you're going to get out. And because of this we can say that the domain is all real numbers because we can put any number we want to infer it. We can put any of these negative x's or any of these positive x's into the function.

But notice how we can also get any negative y's or any positive y's out of this function. So we would see that the range is also all real numbers. So that's the identity function. Now let's take a look at the square function. And the square function forms this interesting shape called a parabola. The parabola is this bowl-like shape that you see on the screen here. Now the square function happens when our function f(x) is equal to \( x^2 \) and the domain for this function is going to be all real numbers. Now, the reason for all real numbers is because notice how all of the negative x's and all of the positive x's are defined by this curve. This curve is continuously expanding up and to the left and right, so we include all of the x values. But what about our range?

We actually don't have all real numbers for the range because, notice even though all of the positive y's are included, we don't include any of these negative y's. The curve does not expand down below the graph. So we can say that our range will go from 0 all the way to positive infinity. And we do include the 0 because we could have a value right here where the origin is, but we can never be below 0 so we can never get a negative output for the square function. Now let's take a look at the cube function.

The cube function says that our function f(x) is equal to \( x^3 \) and notice for this graph that we have all the negative x's and all the positive x's included. So we would say that the domain is all real numbers. So you can get any number you can plug any number you want in for x. And notice how the range for the y values, we include all the negative y's and all the positive y's in this curve. So we can say that the range is also all real numbers.

So that is the domain and range for the cube function. Now let's take a look at the square root function. And the square root function has the most restrictions of all the common functions. So in this function, we have to look at our graph here. Notice how this graph continuously goes to the right and also goes up. And because it continuously goes to the right, we can see we have all positive x's, but notice how none of the negative x's are included.

So we would say that the domain goes from 0, including 0, all the way up to positive infinity. Because we could have our square root function be 0, but we can't ever be below this. We cannot put a negative number in for x Now let's take a look at our range. Notice that our range continuously goes up, so we have all the positive y values, but none of the negative y values are included. So we would say that our range also goes from 0 to positive infinity.

So this is one of the functions where the x values cannot be negative because none of these negative x values are included by this curve. Now let's take a look at the cube root function. The cube root function is where we have our function f(x) equal to the cube root of x. And notice for this function that we have all of our negative x's and all of our positive x's included. So we can say that the domain goes from negative infinity to positive infinity, all real numbers are going to be included for the domain because we see this curve goes to all real numbers. And if we look at our y values, we can see that all of the positive y's and all the negative y's will also be included because this graph continues to go down and up and to the left and right. So all of these numbers will be included as well.

So we can see that the range is going to be all real numbers also. And for the cube root function, our x actually can be negative because we can see that the negative x's are included for this graph. So those are some of the common functions that you'll see throughout this course and also in future math courses. So hopefully you found this helpful and let me know if you have any questions.

Slope-Intercept Form

Hey, everyone. So throughout our discussion on lines, we've seen problems where they ask us to calculate the slope of a line by looking at the rise over the run between two points. But some problems, like the one we're going to work out down below, won't ask for just that. Some problems will ask us to look at the graph of a line and write its equation in a very specific way called the slope-intercept form. So that's what I want to talk about in this video.

And what I'm going to show you is it's basically just a very specific way that we write an equation of 2 variables using 2 things that we've already seen before independently. So we can use its slope and its intercept, specifically the y-intercept. Alright? So I'm going to show you this equation, and it's probably something that you've seen before. So let's get started.

So the slope-intercept form of an equation is actually just an equation. That's y=mx+b. You've probably heard that at some point in a math class, y=mx+b. This is just the slope-intercept form of a line. It's one of the most easy and straightforward ways to describe a line equation.

So let's get started and talk about these two variables here. The m is basically just the slope, and we've already seen that before. We calculate that by using Δy/Δx, rise over run. So in this equation, what we can see is that the rise over the run between these two points is 21, and so the slope is just 2. What about the y-intercept?

Well, the b term over here is the y-intercept, and we've already talked about that separately. The y-intercept is basically just the y-value wherever the graph crosses the y-axis, and it's where x is equal to 0. So for instance, in this case, the graph crosses the y-axis right over here. The y-value at this point is 3. So, that is the b term.

Right? So that's the y-intercept. So to put these things together, y=mx+b, all you have to do is the equation here has a slope of 2, so that goes in front of the x, and then the intercept is 3. So the equation of this line is just y=2x+3. That's how to describe this equation in slope-intercept form.

That's really all there is to it. Alright? So that's all there is to slope-intercept form. Let's go ahead and take a look at another example. Alright?

So in this graph below, we're going to identify the slope and the y-intercept, and then we're going to write the equation in slope-intercept form. So remember, slope-intercept form here is just going to be y=mx+b. In order to figure out m and b, we're going to have to take a look at the graph. Alright? So the easiest sort of value to start with is going to be the b term because you really just have to look at the graph and figure out where it crosses the y-axis.

You don't have to calculate anything. So let's take a look at this graph. Where does it cross the y-axis? Well, it crosses right over here. So, what's the y-value where this graph crosses the y-axis?

It's just negative 3. So that's the b. It's just negative 3. That's all there is to it. You don't have to write the ordered pair.

It's just the y-value. Alright? So how do we calculate m? Well, for m, we're going to have to use rise over run. So we're going to calculate this by using Δy/Δx.

Well, to get from this point to the next one, all I have to do is go up 1 and then over 1. So, up 1 over 1. So that's just a rise over run of just 1. So the slope is 1. I just go 1 over and 1 up. So because the slope is just equal to 1, I basically just plug both of these things into this equation, and my equation is just equal to y=x-3. So this is the equation that describes this line in slope-intercept form, y=x-3.

Alright? So, anyway, folks, thanks for watching. Let me know if you have any questions.

In the graph shown, identify the y–intercept & slope. Write the equation of this line in Slope-Intercept form.



• A. $y=\frac23x+1$
• B. $y=-\frac23x+1$
• C. $y=-2x+1$
• D. $y=x+2$

Properties of Parabolas

Hey everyone, we've worked with some basic graphs of functions and here we're going to focus on one specific type of function, the quadratic function. Now, it might seem like a lot of information at first and this graph behind me might look a little bit intimidating, but don't worry, I'm going to walk you through absolutely everything that you need to know about the graphs of quadratic functions here and you'll be an expert in no time. So let's go ahead and jump right in. A quadratic function is a polynomial of degree 2 that has a standard form of f(x) = ax² + bx + c. Now, all of these functions here are examples of quadratic functions and you can see that a, b, and c can be any real number whether it be a fraction, a negative number, or even 0 so long as a is not 0 and our largest exponent is still 2, making it a quadratic function.

Now here we're going to focus on the graphs of quadratic functions so let's go ahead and take a look here. We've worked with the square function before f(x) = x² and you may remember that this square function was a parabola. Now this is actually going to be the same for all quadratic functions. They are all going to have this curved parabolic shape. Whether it is right-side-up, upside-down, or located anywhere on our coordinate plane, our quadratic functions will always be that same shape of a parabola.

So we're going to look at the different elements of a parabola here and some are going to be ones that we're familiar with like the x intercept or the y intercept but we're also going to work with some that are specific to parabolas, like the vertex or the axis of symmetry. So let's go ahead and get started with our vertex. The vertex of a parabola is either going to be the lowest point or the highest point depending on whether our parabola is opening upwards or downwards. So for our square function here, we see that our vertex is right at that origin point and we're always going to write our vertex as an ordered pair, so my vertex is simply (0, 0). Now, for my other function here, it is not at the origin, it is actually at the point (-2, 1), so my vertex is (-2, 1).

Now the other thing that I want to consider with my vertex is whether I'm dealing with a minimum or a maximum point. So again, looking at my square function here, this is the lowest point on my graph of that quadratic function. So it is a minimum point. It's all the way at the bottom and nothing goes below it on that parabola. So not only is my vertex here (0, 0), it also represents a minimum.

Now for my other function, my vertex is all the way at the top, which tells me that I am not dealing with a minimum. I'm actually dealing with a maximum point here. So here, my vertex is (-2, 1), and it is a maximum. So let's go ahead and look at something that we are more familiar with, our x intercept. Now, the x intercept of a graph is anywhere that our graph crosses our x-axis, not our y-axis, our x-axis.

So we're going to look for the points that cross that x-axis. And between my two functions here, I have 3 different points that cross my x-axis that represent my x intercepts. So let's take a look at our square function. Now, there's only one point that crosses the x-axis and it is at my origin, so my x intercept is simply 0. Now looking at my other function, there are 2 points that cross the x-axis, so I need to account for both of them, (-3, 0) and (-1, 0), as my x intercepts.

Now you'll only ever have 1 or 2 x intercepts, never less or more than that. So let's move on to our y intercept. Now the y intercept is where a graph crosses the y-axis, this time the y-axis. So let's look at our square function first. We have this point.

This represents our y intercept. Now our y intercept here is again at the origin, simply at 0. And for my other graph, my other function, my y intercept is down here at -3, so it's simply -3. Now we've looked at our intercepts and our vertex. Now we want to look at our axis of symmetry.

And now our axis of symmetry is something that is going to be specific to parabolas, and it represents the line that divides our parabola perfectly in half. It is symmetric about that line. So our axis of symmetry is going to perfectly cut it in half. So looking at my square function, it's going to go straight through the middle and it's actually always going to go straight through our vertex. So here, my axis of symmetry is simply the line x = 0, a vertical line through my vertex.

Now looking at our other function here, if I draw a line that divides this perfectly in half, you're often going to see this actually written with a dotted line but I've just highlighted them here so that it's easier to see. But here again, we're going straight through that vertex point at x = -2. So it's always going to be a straight vertical line through our vertex and that represents our axis of symmetry. If I were to fold my parabola in half at that line, it's going to perfectly match up because it is symmetric about that line. Now something else to consider whenever we're looking at graphs of functions is always the domain and the range.

So first looking at our domain, the domain of all quadratic functions is actually always going to be the same and it's always going to be negative infinity to infinity, which you might also recognize as being all real numbers. So for both my square function and for my other function here, they are both all real numbers, negative infinity to infinity. Now our range is going to depend on whether we have a minimum or a maximum point. So looking first at our square function, we know that we're dealing with a minimum point here and that looking at that, my y values can go anywhere from that minimum and all the way up beyond that. So whenever I'm dealing with a minimum, my range is always going to go from y minimum, whatever that minimum point is, all the way up to infinity.

So in this case, since I start down at 0, my range is simply 0 to infinity. Now when I'm dealing with a maximum point like I have here on my other graph, it's always going to go from my maximum point down to negative infinity. So in interval notation, I would write that as negative infinity up to my maximum point, y maximum. So in this case, it goes from negative infinity up to my y point at y = 1, and that's my domain and range. Now we want to look at one last thing on our graphs here.

Something that you're going to often be asked when dealing with parabolas is to tell them the interval for which your parabola is increasing and decreasing. Now this is just a fancy way of saying for what values of x is our graph going up and for what values of x is it going down. So let's just take a look at our second function here that's facing downward and decide where it's increasing and decreasing. Now looking at the first side of this, going down here all the way up to my vertex point, we see that our graph is going up. It is increasing.

So the interval for which it is increasing is simply from negative infinity up to that vertex point at -2. And our vertex is actually always going to divide the intervals for which we're increasing or decreasing. So the other side, from -2 all the way to infinity, we're going to be decreasing because we see our graph is going down. So this is just simply the opposite, -2 to infinity. So that's all the stuff we need to know about parabolas.

One more thing that I want to mention here is that you might have noticed that my equation here doesn't quite look like it's in standard form. And that's actually because we're often going to see quadratic functions written in what's called a vertex form, which is what that is right there. And it's actually going to help us to graph with ease. It's going to help us to easily graph these quadratic functions, which is what is coming up next. So thanks for watching.

I'll see you in the next video.

Identify the ordered pair of the vertex of the parabola. State whether it is a minimum or maximum.

• A. $\left(-3,-2\right);$ maximum
• B. $\left(-3,-2\right);$ minimum
• C. $\left(-1,2\right);$ maximum
• D. $\left(-1,2\right);$ minimum

Where is the axis of symmetry located on the given parabola? 

• A. $x=3$
• B. $x=0$
• C. $x=1$
• D. $x=4$

Introduction to Polynomial Functions

Hey, everyone. We've worked with polynomial expressions, and we've even worked with a specific type of polynomial function, a quadratic function, where my highest power is 2. Now I want to take a broader look at all the possibilities of polynomial functions, which can really just be any polynomial, but now f(x) is equal to that polynomial, making it a polynomial function. That might sound like a lot, any polynomial, and now it can be a function too. But don't worry.

We're going to rely on a lot of what we already know about polynomials and some of what we just learned about quadratic functions in order to learn more about polynomial functions and their graphs. So let's go ahead and jump right in. Looking at the polynomial function that I have right here, we want to remind ourselves of a couple of things that we learned with polynomials. The first of which is that polynomials can only have positive whole number exponents. So that means no negatives and no fractions in those exponents.

The other thing is that whenever we write our polynomials in standard form, all of our like terms need to be combined and it needs to be written in descending order of power. So if I start with a power of 3, my next power is going to be 1 lower and then 1 lower, 1 lower until I get to the last one in descending order. Now looking at these polynomials that I have here, you might notice this one looks similar to what you've seen in your textbook. This can look a little bit intimidating when you first see it, but don't worry. This is just showing us exactly how to write any polynomial in standard form.

So this \( a_n \) right here represents my leading coefficient. So in the polynomial that I have here, it's simply 6 and \( n \) represents the degree of the polynomial. So in my polynomial here, it's just 3. Now when I look at my next term, I see that this power goes down to \( n-1 \), and that's just showing us that it's in descending order of power. So here I started with 3.

If I subtract 1 from that, I get 2, which is my very next power here. And then all of these \( a \) terms down to my \( a_0 \) just represent all of my coefficients and then my constant. So don't let that form freak you out; it's just showing you how to write in standard form. Okay, so let's just look at some polynomial functions here. So looking at this first one, I have \( f(x) = -x^2 + 5x^3 - 6x + 4 \) and I want to determine if this even is a polynomial function and if it is, write it in standard form and state both the degree and leading coefficient.

So let's first determine if this even is a polynomial function. And since I only have positive whole number exponents, I don't have any fractions or any negatives in those exponents, it looks like, yes, this is a polynomial function. So let's go ahead and write it in standard form. Now here, it looks like I just need to switch these two terms in order for it to be in standard form. So I end up with \( f(x) = 5x^3 - x^2 - 6x + 4 \), and we're good.

We're in standard form. So let's go ahead and identify that degree. So the degree of this polynomial, looking at that first term, my highest power, is simply 3, and then my leading coefficient is the one that's attached to that \( x^3 \). It is 5. So we're good on that one.

Let's go ahead and move to our next one. Here we have \( f(x) = 2x^{1/2} + 3 \). Now is this a polynomial function? Well, I have a fraction in my exponent there, so no, this is not a polynomial function. I cannot have fractions as exponents, so I don't have to worry about anything else because it's not a polynomial function at all.

Let's go ahead and move on to our last example here. We have \( f(x) = -\frac{2}{3} x^4 + 1 + 3 \). Now looking at this, there is a fraction as a coefficient. Does that mean that it's not a polynomial? Well, no, because it's just a coefficient.

It's not in my exponent and my exponents here are positive whole numbers, so it looks like yes, this is a polynomial function. You can have a fraction as a coefficient, just not as an exponent. So let's go ahead and write this in standard form, which I can do by simply combining these like terms that I have here. So I end up with \( f(x) = -\frac{2}{3} x^4 + 4 \), combining that 1 and 3. So let's identify our degree here.

So my highest power on that first term

Determine if the given function is a polynomial function. If so, write in standard form, then state the degree and leading coefficient. $f\left(x\right)=4x^3+\frac12x^{-1}-2x+1$

• A. Polynomial with $n=3,a_{n}=4$
• B. Polynomial with $n=4,a_{n}=3$
• C. Polynomial with $n=-1,a_{n}=\frac12$
• D. Not a polynomial function.

Determine if the given function is a polynomial function. If so, write in standard form, then state the degree and leading coefficient. $f\left(x\right)=2+x$

• A. Polynomial with $n=1,a_{n}=2$
• B. Polynomial with $n=0,a_{n}=1$
• C. Polynomial with $n=1,a_{n}=1$
• D. Not a polynomial function.

Determine if the given function is a polynomial function. If so, write in standard form, then state the degree and leading coefficient. $f\left(x\right)=3x^2+5x+2$

• A. Polynomial with $n=3,a_{n}=2$
• B. Polynomial with $n=2,a_{n}=3$
• C. Polynomial with $n=2,a_{n}=2$
• D. Not a polynomial function.

Intro to Rational Functions

Hey, everyone. By now we've worked with a bunch of different polynomial functions, things like \( x^2 + 4x + 1 \) or \( 3x + 2 \) or other polynomials of different degrees. But if we take these polynomials and we put one on the top of a fraction and the other on the bottom of a fraction, we now have a new function called a rational function that has a polynomial of any degree in both the numerator and the denominator of a fraction. Now you might see this written in your textbooks as \( p(x) \) divided by \( q(x) \), where \( p(x) \) just represents any polynomial and \( q(x) \) is any other polynomial. Now I know that fractions might not be your favorite thing to work with, and taking polynomials and putting them into fractions might seem a little bit intimidating.

But we've worked with polynomial functions before, and we've even already worked with rational equations. So in working with rational functions, we're going to rely on rational equations. So in working with rational functions, we're going to rely on a lot of what we already know about these functions and equations that we've already worked with. And I'm going to walk you through everything that you need to know step by step. So let's go ahead and get started.

Now when working with rational equations, we said that our denominator could not be equal to 0. So for this particular equation, looking at my denominator, I would say that \( x \) cannot be equal to 1, and that would be referred to as my restriction because 1 would make my denominator 0, which is not something that I want to happen. Now the same thing is true of rational functions because the same thing is true of all fractions. The denominator can never be 0. Now because we're working with a function now that can be evaluated for any value of \( x \), we're going to refer to this as our domain restriction.

So it's the same exact idea. Our denominator cannot be 0. We're now just referring to our entire domain and defining what \( x \) can and can't be based on that restriction. So looking at this function that I have here, \( \frac{1}{x-1} \), I have the same restriction. But writing it in my domain, I write my domain in set notation in \( x \) such that \( x \) cannot be equal to, in this case, 1 because that's what would make my denominator 0.

Now we write it like this because we determine our domain the same exact way. We simply set our denominator equal to 0 and solve for \( x \) to get that restriction. And then we know that our domain can be any real number besides that number that makes our denominator 0. So now that we know how to find the domain, let's go ahead and find the domain of a couple of different rational functions here. Now looking at this first rational function, I have \( \frac{3}{3x-12} \).

So we're simply going to take our denominator and set it equal to 0, solving for \( x \). Subtracting 12 from both sides, I end up with \( 3x = -12 \). And then to isolate \( x \), I want to go ahead and divide both sides by 3. Now that leaves me with \( x = -4 \), and that tells me my domain restriction. I know that \( x \) can be any real number, except \( x \) cannot be equal to negative 4 because if I took negative 4 and I plugged it back into my denominator, it would make it 0, which is what I don't want.

So my domain here is \( x \) such that \( x \) cannot be equal to negative 4. Now let's look at another function here. So I have \( f(x) = \frac{x+5}{x^2-25} \). So again, taking our denominator, setting it equal to 0, if I add 25 to both sides, canceling out that 25, leaving me with \( x^2 = 25 \). And to isolate \( x \), I simply take the square root of both sides, leaving me with \( x = \pm \sqrt{25} \), which I know is just 5.

Now with this answer, this tells me that my domain is actually restricted by 2 values. It can't be 5 or negative 5. So my domain is \( x \) such that \( x \) cannot be equal to 5 or negative 5. We can have multiple domain restrictions depending on what our denominator is. Now that we know how to find the domain of different functions, something else that you'll be commonly asked to do when working with rational functions is to write them in lowest terms.

Now, in order to write functions in lowest terms, we're going to factor something that we've done a million times before. We're going to factor both the top and the bottom of our function and then simply cancel out any common factors. So we're just writing our function in the simplest terms, simplifying it as fully as we can. So let's take another look at these functions here. So starting again with \( \frac{3}{3x+12} \), let's go ahead and factor the top and the bottom here.

Now, I can't really factor the top because it's just the constant. So I'm simply left with 3 on the top there. But on the bottom, it looks like I can pull out a greatest common factor of 3 from both of these terms, leaving me with \( 3(x+4) \). Now we want to go ahead and cancel out any common factors. In this case, I have a common factor in my numerator and my denominator of 3, so I can go ahead and cancel that out, leaving me with my function \( \frac{1}{x+4} \) as all that's left.

So my function in lowest terms here is simply \( \frac{1}{x+4} \). Now let's look at our other function. So we have \( f(x) = \frac{x+5}{x^2-25} \). So let's go ahead and factor. Now again, my numerator can't be factored, it's already just \( x+5 \), but my denominator is a difference of squares so this factors to \( (x+5)(x-5) \).

Now I want to go ahead and cancel our common factors. In this case, I have a common factor of \( x+5 \) that I can cancel from both my numerator and my denominator. So canceling that factor out leaves me with \( \frac{1}{x-5} \) and that's my function in lowest terms. Now something that I want to point out here is that if I were to take this function in lowest terms and solve for the denominator here by setting my or solve for the domain here, setting my denominator equal to 0, I would only get one of the values that appeared in my domain restriction up here. I would say that \( x \) cannot be equal to 5, but I wouldn't know about that negative 5.

That's because you always need to find your domain before you write in lowest terms. Never write in lowest terms and then find the domain because you might be leaving something out. Okay. Now that we know the basics of working with rational functions, let's move on. I'll see you in the next video.

Find the domain of the rational function. Then, write it in lowest terms.

$f\left(x\right)=\frac{x^2+9}{x-3}$

• A. {x∣x≠0}, $f\left(x\right)=\frac{1}{x-3}$
• B. {x∣x≠3}, $f\left(x\right)=\frac{x^2+9}{x-3}$
• C. {x∣x≠−3}, $f\left(x\right)=\frac{x^2+9}{x-3}$
• D. {x∣x≠3}, $f\left(x\right)=x+3$

Find the domain of the rational function. Then, write it in lowest terms.

$f\left(x\right)=\frac{6x^5}{2x^2-8}$

• A. ${\left\lbrace x|x\ne2,-2\right\rbrace},f(x)=\frac{3x^5}{x^2-4}$
• B. ${\left\lbrace x|x\ne2,-2\right\rbrace},f(x)=\frac{6x^5}{2x^2-8}$
• C. ${\left\lbrace x|x\ne2\right\rbrace},f(x)=\frac{3x^5}{x^2-4}$
• D. ${\left\lbrace x|x\ne2\right\rbrace},f(x)=\frac{3x^5}{x^2-8}$

Exponential Functions

Welcome back, everyone. By now, we've worked with a ton of different polynomial functions, and one of the most common and perhaps most basic polynomial functions we've seen is f(x)=x2. But if I take that x and that 2 and I swap them, I now have f(x)=2x, which is an entirely different type of function called an exponential function that we're going to have to evaluate and graph just as we have with other functions. Now, I know seeing that variable in the exponent might seem a little bit strange at first, but here I'm going to walk you through the basics of exponential functions using a lot of what we already know about exponents themselves and what we know from working with other functions. So let's go ahead and get started.

Now looking at our function here, 2x, we have two different things going on. We have this base number of 2, and then we have the power that that base 2 is raised to, in this case, x. Now, in working with exponential functions, we need to consider a couple of different things about the base of our function. So in this case, we have 2. But the base of any exponential function needs to be a constant, so it can't be something with a variable that changes. It has to be positive. So we can't have any negative numbers as our base. And it also cannot be equal to 1. So something like 2 fits all of that criteria. Now when considering the exponent or the power of our exponential function, we only need to consider one thing, and that is that it contains a variable.

So it's something that can change, something like x. Now that we know a little bit about exponential functions, let's take a look at a couple of different exponential functions down here. So here we have this function f(x)=23x. And here we want to determine if this even is an exponential function. And if it is, then we can go ahead and identify both the power and the base.

So looking at this function here, I have 23 as that base. And I want to consider my three things: constant, positive, and not equal to 1. So 23, it is constant. It is positive. And it is not 1.

So it looks like we're good so far. Now looking at our power here, we have this power of x. We need to make sure our power contains a variable, which it already is a variable just by itself. So it looks like we are good here, and this is an exponential function. Now to identify our power and our base, we saw our power of x there, so our power is simply x by itself.

And then our base is 23. Now it's okay to have a base that's a fraction as long as it's constant and positive. Let's move on to our next example. So here we have f(y)=1y. Now our base here is 1.

1 is constant. It is positive, but it is 1, so it is not meeting that last criterion not being equal to 1. So I don't even have to look any furt...

Determine if the function is an exponential function.  

If so, identify the power & base, then evaluate for $x=4$.

$f\left(x\right)=\left(-2\right)^{x}$

• A. Exponential function, $f\left(4\right)=16$
• B. Exponential function, $f\left(4\right)=-16$
• C. Not an exponential function

Determine if the function is an exponential function.  

If so, identify the power & base, then evaluate for $x=4$ .

$f\left(x\right)=3\left(1.5\right)^{x}$

• A. Exponential function, $f\left(4\right)=410.06$
• B. Exponential function, $f\left(4\right)=15.19$
• C. Not an exponential function

Determine if the function is an exponential function.  

If so, identify the power & base, then evaluate for $x=4$ .

$f\left(x\right)=\left(\frac12\right)^{x}$

• A. Exponential function, $f\left(4\right)=\frac{1}{16}$
• B. Exponential function, $f\left(4\right)=-16$
• C. Not an exponential function

Logarithms Introduction

Hey, everyone. Whenever we solve polynomial equations like x3=216, we can simply do the reverse operation in order to isolate x. So here, since x is being cubed, I could simply take the cube root of both sides to cancel that, isolating x, giving me my answer that x=216. But what if my variable x is instead in my exponent, like this 2x=8? How are we going to isolate x here?

Well, I know that looking at this, I can just think, okay, how many times does 2 need to be multiplied by itself to give me 8? And I know that my answer is simply x is equal to 3. But what if I'm given something like 2x=216? I really don't want to have to multiply 2 enough times to get up to 216. So is there not just an operation I can do to cancel out that 2 and leave me with x?

Well, here I'm going to show you that there is an operation that does just that because the reverse or inverse of an exponential is actually taking the logarithm or a log. Now, the first time you work with logs, they can be a bit overwhelming. But here I'm going to walk you through exactly what a log is and how we can use it to actually make our lives easier, especially when working with exponents. So now that we know that the reverse or inverse operation of an exponential is simply taking the log, we can go ahead and take the log of both sides of this equation in order to isolate x. So I take log 2(2x), and that's equal to log(216).

But we actually do need to consider one more thing here because whenever we cancelled the 3 on our x3, we took the cube root. We didn't take the square root or the 4th root or anything else. We took the cube root to cancel that 3. And we need to consider something similar when working with logs because logs and exponentials need to have the same base as each other to cancel. So here, since I have an exponential of base 2, I want my log to have a base of 2 as well.

So really, I want to take a log base 2 of 2x, and then on the other side, a log base 2 of 216. Now that my log and my exponential have the same base, then it's going to fully cancel out, leaving me with just x=log 2(216). Now it's fine to leave it in this form here. This is actually called our logarithmic form. And we're later going to learn how to fully evaluate these and get a number, but for now, we're just going to keep it in that log form.

Now that we're here, what exactly does this statement mean? We have log 2(216). Well, a log is actually giving us the power that some base must be raised to in order to equal a particular number. But what does all of that mean? Well, looking at our function here or our equation here, log 2(216), this is really saying, okay, what power does 2 need to be raised to in order to give me 216?

Now, this statement here I mentioned is in its logarithmic form, and it's actually an equivalent statement to our very first equation, 2x=216. This is just in its exponential form, and we basically translated it into its logarithmic form. Now we're going to have to do this for multiple statements. We're going to have to translate and convert expressions between these two forms. So diving a bit deeper in converting between these two forms, let's start by taking this equation in its exponential form, 3x=81, and putting it into log form.

Now whenever we convert between these two forms, no matter what we're going to or from, we're always going to start at the same place. We're going to start with our base. So this base 3 of our exponent is going to become the base of our log.

Graphs of Logarithmic Functions

Hey, everyone. Earlier when graphing exponential functions, we found and plotted a bunch of different ordered pairs, and we found that our graph had similarities to both polynomial and rational functions as we approached an asymptote. Now we're faced with graphing another type of function, a logarithmic function, and you may be dreading having to learn how to graph yet another different type of function. But I have great news for you here because graphing a log function is actually super similar to graphing an exponential function because a log is the inverse of an exponential. So here, I'm going to walk you through graphing a log function using exclusively things we already know about graphing exponential functions.

So let's go ahead and get started. Now looking at the log function we have here, we have a log base 2 of x. So let's start by just plugging in some values of x and getting some ordered pairs. So starting with x equals 1, if I plug that into my function, I get log base 2 of 1. And I know any time I take the log of 1, no matter what the base is, I'm always going to get 0.

So I have my first ordered pair at (1,0), which I can go ahead and plot on my graph. Then if I plug in x equals 2 to my function, I get log_2(2), which since my base is the exact same as what I'm taking the log of, I know that my answer here is going to be 1, and I get my 2nd ordered pair at (2,1). Now let's pause here and take a closer look at these two points. I have 1,0 and 2,1. Now looking over to my exponential function here, I have f(x)=2x, which is the inverse function of this log function that we are trying to graph here.

Now looking at my exponential function, I have these two points, 0,1 and 1,2. And looking back to the points that we just found, I noticed that these are the same exact points with x and y flipped. Now this is because these are inverse functions, so this is actually going to work for every single x and y value of our exponential function. If we simply flip them, we'll have all of the new ordered pairs needed to graph our log function. So let's go ahead and flip our x and y values and get these points plotted on our graph.

So looking at our first point here, -2,14 and having flipped those points, there's my new point. And then for -1,12. Now flipping these final two points here, I'll end up with 4,2 and 8,3. And we can go ahead and plot all of these on our graph. So 4,2, 8,3, and then for our small fractions, -1,12 and -2,114.

Now connecting all of these points, I have a complete graph of my log function. And I notice here on this side that we're getting really close to that y-axis but not quite touching it, which tells us that we're, of course, dealing with an asymptote that I can go ahead and plot using a dashed line right along that y-axis to form my vertical asymptote. Now that we have the complete picture of the graph of our log function, let's compare it to our original exponential function. Now looking at the graph of my exponential function here, if I were to take my entire graph and fold it along this diagonal line and then stamp my exponential function onto the other side, I would end up with a graph identical to the log function that we just graphed because this is actually exactly what we did. We simply reflected the graph of our exponential function over this diagonal line, y equals x, to get the graph of our log function.

Now this will work for any log function and its corresponding inverse exponential function because they're inverses. They're always going to be reflections of each other along the line y equals x. Now if you want a way to remember the shapes of these graphs, remember when you're dealing with an exponential function, you can think of a lowercase e. And if we just extend the tail of that e out, it looks really similar to the shape of the graph of our exponential function. And for a logarithmic function, if we look at our word logarithmic and we take that lowercase r and extend that r out, we get a shape really similar to the graph of our logarithmic function.

So remember, for an exponential function graph, the shape looks really similar to extending a lowercase e, whereas for a logarithmic function, we extend that lowercase r. Now that we have a good idea of what's going on with the shapes of these graphs and what they look like, let's also consider their domain and range. Now when graphing our log function, we were able to flip the x and y values of the inverse exponential function, and we can actually do the same exact thing to our domain and range. So the domain of our exponential function corresponds to the range of our inverse log function. So it's going to be all real numbers for any log function of any base.

Then also, the range of our exponential function corresponds to the domain of our exponential function corresponds to the domain of our log function, and it's going to depend on our asymptote and in this case go from 0 to infinity. Now speaking of asymptotes, we also can take a look at our asymptote here, and we had a horizontal asymptote at y equals 0 for our exponential function, which then corresponds to a vertical asymptote now at x equals 0 for our logarithmic function. Now let's consider one final thing here and consider when we have to graph logs of different bases. Now everything that we've looked at so far has been kind of the opposite of working with an exponential function but this thing is going to be the exact same and the direction of our graph is going to depend on the value of our base b just as it did for exponential functions and it's going to depend on it in the exact same way. So for values of b that are greater than 1, like 2 or 3 or 10 or whatever, our graph is going to be increasing.

And for values of b between 0 and 1, like, say, 12, our graph is instead going to be decreasing. Now that we know everything that we need to about graphing log functions, let's get some more practice.

Graphing Logarithmic Functions Example 1

Hey everyone. In this problem, we're asked to graph 2 different functions, \( f(x) = 3^x \) and \( g(x) = \log_3(x) \). Now we can see that these functions are inverses of each other because the logarithm is the inverse function of an exponential, and we're going to go ahead and graph both of them together. So let's start with \( f(x) = 3^x \) and find and plot some points here. Starting with \( x = 0 \), we can plug that into our function and get \( 3^0 \), which is just 1.

Now if I plug 1 into my function, that's \( 3^1 \), which is just 3. And then \( 3^2 \) is going to give me 9. Now let's go ahead and plot these points on our graph. So I'm going to start with 1, and then I have 3, 9 is completely off of my graph, actually.

We know that it's going to increase really rapidly and get really steep there. Now for my negative values of x, we're going to see the same numbers but instead fractions as we had with positive 1, 2. So \( 3^{-1} \) is simply \( \frac{1}{3} \) and then \( 3^{-2} \) is \( \frac{1}{9} \) and we can see that we're getting smaller and smaller as we go into those negative numbers. So let's go ahead and plot those points at negative one, \( \frac{1}{3} \), and then 2, \( \frac{1}{9} \), really close to my x-axis. So we can go ahead and connect all of these points that we have on our graph here.

I know that I'm going to get really steep on this side. And then on my left side here, I'm going to get really close to my x-axis but not quite cross it because we have an asymptote there. So I can go ahead and plot my asymptote as well. I know that it's at the x axis right at \( y = 0 \). Now that we have our graph for \( 3^x \), let's move on to plotting \( g(x) = \log_3(x) \).

Now remember that whenever we're dealing with inverse functions, we can simply swap our x and y values. So these values that I have for x over here are simply going to become my y values when working with my new function that is the inverse. And then my y values over here are going to become the x values of my inverse function. So let's go ahead and swap all of those points so that we can get them plotted on our graph. So starting with this point, I'm going to go ahead and switch that up.

So now it is 1,9 negative 2, just reversing my x and y. And then for negative one, one third, that becomes one third a negative one. 0,1 flip is 1,0. Then I have 3,1, which I flipped to get 1,3. And then my last point, flipping that 2 and that 9, I get 9,2.

Now let's go ahead and plot all of these points for our function \( g(x) \) to get our final function up here. So let's start with this 1,0 and work our way down this way first. So 1,0 and then I have 1,3. And then 9,2 goes off of my graph yet again. And then going the opposite direction into my fractions, one third negative 1 is going to be right about here.

And then one ninth negative 2, we're going to be getting really close to our y-axis this time. So we can go ahead and connect all of these points to form our graph. And then we're getting really close to that y-axis, which tells us that we're dealing with a vertical asymptote up on this side right on our y-axis at \( x = 0 \). So now that we have plotted both of our functions, we can go ahead and determine our domain and our range. Now when dealing with our domain of our first function, \( f(x) = 3^x \), I can see that this is going to be any value of x.

It can be all real numbers. Now, we know that our domain of our \( 3^x \) is going to match up with the range of our function \( \log_3(x) \). So I know that my range here is also going to be all real numbers, which I can verify by taking a look at my function here. Now, for my range of my function \( 3^x \), I need to go ahead and take a closer look and identify where my asymptote is so that I can get an accurate range here. Now, when I consider my range, I know that we're not going to go past that asymptote here.

So my range is simply going to go from my asymptote at that 0 point all the way to infinity. So my range here is simply from 0 to infinity. Now we know that our range is going to flip and become the domain of our other function of our inverse function, so that tells me that the domain of \( \log_3(x) \)is going to be from 0 to infinity, which we can again verify by taking a look at our graph. We see that we can't cross this asymptote, so we start at that 0 point and go to everything positive. Now that we've graphed these inverse functions, let's get some more practice.

Let me know if you have any questions.