Today we're going to focus on graphing linear equations, including those with inequalities, quadratic equations, we're going to talk about transformations, graphing radical functions, cubic functions, absolute value equations, rational expressions, exponential, and even logarithmic equations. So let's begin, let's start with the basics. Let's say if we want to graph y is equal to 2x plus 3. Now there's many ways to graph it.
So the first way we're going to use is, we're going to make a table. Let's keep things simple. So let's say if you have your x, y table, let's plot some points. So we're going to plot the point negative 1, 0, and 1. We don't really need that many points. Okay, so at negative 1, if we plug in negative 1 for x, so 2 times negative 1 plus 3, that's negative 2 plus 3, that's 1. If you plug in 0, you should get 3. If you plug in 1, 2 plus 3 is 5. And then you just got to plot those points.
So here's negative 1 on the x-axis. And over here, y is 1. That's our first point. When x is 0 and y is 3, we have this point here.
I'm going to plot another point because my graph doesn't go up to 5. So I'm going to plot negative 2. If we plug in negative 2 for x, that's 2 times negative 2, that's negative 4. Plus 3, we get negative 1. So that negative 2 is around somewhere over here. And then you could just connect those points with a line. This is just a rough sketch. My graph is not that perfect, but that's the general idea of how to plot or how to graph a linear equation using a table. You just got to pick a few points and just plug it in.
But now let's see if we can graph another question just like that, but without making a table. Let's say if we have y is equal to negative 2x plus 1. Okay, this equation is in slope-intercept form, mx plus b. b is 1, that's your y-intercept.
So that's where your graph is going to start, at 1. m is the slope. The slope is negative 2. which is the same as negative 2 over 1. This is your rise, and this is the run. The rise over run.
So every time you move one unit to the right, you need to go down two units. So one unit to the right, down two. Our next point. is over here and if we go money to write down to here's an exploit and once units are right down to you it's gonna be somewhere over here and then we just gotta connect those points for here So that's the slope-intercept method.
You just plot the y-intercept and then use the slope to find each successive point. So now let's try another example just like that, but using fractions. So let's say if you have this equation, y is equal to negative 3 fourths x plus 5. so one two three four five one two three four let's go up to 85678 so the winder step is five so that's going to be the first point now the slope is negative 3 over 4 so negative 3 is the rice and four is the run so we're going to travel for units to the right and we're gonna go down three units so our next point is over here and then we're gonna do the same thing we're gonna travel four units to the right towards eight and down three so we're gonna be down at negative one and then just connect those points with a line So go ahead and try this example.
Feel free to pause the video and give this problem a shot. And when you're ready, just unpause it. Okay, so let's begin. Let's begin by plotting the y-intercept, which is at negative 2. And the slope is 2 over 3, so our rise is 2, and the run is 3. We're going to go up 2 units, and then over 3 units to the right. And then same thing, up 2, over 3. So we got these three points to plot.
And then once you feel like you have enough points, you really only need at least two to make a good line. But once you have enough points, you can just plot it. So that's how you can graph linear equations. in slope and a step form.
But now let's talk about how to graph it when it's in standard form. Standard form is ax plus by is equal to c. So let's say if we have this equation, 2x plus 3y is equal to 6. Now what you want to do is you want to find the x and y intercepts.
So to find the x intercept, plug in 0 for y. So this becomes 0. So 2x equals 6. If you divide both sides by 2, you get x equals 3. So you get the point 3, 0. So that's the x intercept. To find the y intercept, plug in 0 for x.
So you get 3y is equal to 6, and if you divide both sides by 3, y is equal to 2. So your y-intercept is 0, 2. And that's what you want to do if you have an equation in standard form. And then once you have those two intercepts, just plot those two points and connect them with a line. So 1, 2, 3, 4. 1, 2, 3, 4. Okay, so the first point is at 0, 2. The next one is at 3, 0. And then just connect those two points with a line.
That's the simplest way to graph an equation in standard form. So let's try another example. Let's say if we have... 3x minus 2y is equal to 12. Go ahead and try this, and then plot it when you're ready. so let's find the x-intercept if we plug in 0 for y the 2y disappears so we're just going to get 3x is equal to 12 and if we divide both sides by 4 12 divided by 3 12 divided by 3 is 4 so we get the point four comma zero and for the y-nose up we're going to plug in zero for X so we're just going to get negative 2y is equal to 12 and we're going to divide both sides by negative 2 so we're going to get y is equal to negative 6 excuse me which is the point 0 negative 6 so we have 0 negative 6 and 4 0 and we're just going to connect those two points with a line I don't have a ruler so that's just my graph won't look perfectly straight but you get that the point though so now you know how to plot it in standard form how would you graph an equation that looks like this Y is equal to 3 how would you plot that All you got to do, if y equals a number, it's just a horizontal line at 3. That's how you graph it.
And if you get a question that looks like this, let's say x is equal to 2, it's simply a vertical line at 2. So it looks like this. So that's how you can graph a linear equation if y or x equals a constant. But now let's talk about graphing inequalities. Let's say if you have this question.
y is greater than 2x minus 4, and y is less than or equal to negative 3 over 2x plus 3. By the way, if it's in standard form, make sure to convert it to slope intercept form. Which is what we see here. Keep in mind, slope and step form is like this. Y equals mx plus b.
You want y to be on one side of the inequality. This is going to make it a lot easier to graph in. Okay, so go ahead and plot these equations in the meantime. so let's plot the first one so the y intercept is negative 4 and the slope is 2 so every time we move one unit to the right we're gonna go up to one unit to the right up to the slope is 2 over 1 And notice that y is greater than 2x minus 4, but not greater than or equal to.
If it's just greater than, we need to use a dashed line. Now for this one, this is going to be a solid line because it's less than or equal to negative 3 over 2x plus 3. But for this one, make sure you use a dashed line. so now let's graph the other equation so the y-axis is at 3 and the slope is negative 3 over 2 that means as you move two units to the right the run is 2 the rise is negative 3 so we need to go down 3 so over 2 down 3 and then over 2 down 3 again and then over 2 down 3 And it's going to be a solid line. Okay, so now we need to know which region to shade. There's four regions in the graph.
This is the first one, this is the second one, this is the third, and this is the fourth. Ignore the x and y axis. Look at the four regions created by the green and yellow line. When you shade it, it's going to be one of those four regions.
So let's focus on this equation first. y is greater than 2x minus 4. So the green line is the 2x-4 line. Greater than, I'm going to shade it in white, is anywhere above that line. So that entire region. that's greater than the line than the green line now for the yellow line y is less than or equal to so it's below the yellow line i'm going to shade that in purple so this is the region below the yellow line Now, we need to shade the region where it's true for both.
And it's true in this region here. That's where we have both the purple and the white line. So this is where the answer is. On the test, this is the only region you should shade. The other ones, erase the marks that you made.
And that's how you choose the right region to shade. So let's try another example. Let's say if x is greater than negative 3, y is less than or equal to 5, and y is greater than x minus 3. So, x is at negative 3. That's going to be a vertical line.
And it's going to be a dashed line. Because it's just greater than, but not equal to. And y is less than 5, so that's a horizontal line at 5, but it's a solid line. y is greater than x minus 3 so the slope intercept I mean the intercept the y intercept is at negative 3 and there's a 1 in front of X so the slope is 1 so every time you move 1 to the right you go up 1 and it's going to be a dash line now need to know where to shade it so let's look at this first equation X is greater than negative 3 it's greater than negative 3 to the right anywhere to the right of this function now Y is less than 5 it's less than 5 below this line anywhere below this line and why is also greater than x-3 it's greater than x-3 above this line so we need to choose the region in which all arrows are located in if you notice it's true within this triangle so this is the reason that we shade because all three equations are true in that region and that's how you do it Okay, so now let's move on to quadratic equations. So let's say if you have this graph, y is equal to x squared plus 2x plus 3. Notice that it's in standard form.
ax squared plus bx plus C Now for this example we're going to make a data table. We're going to plot points so if you want to make a data table you want to find the vertex first to make your life a lot easier and The x-corner of the vertex is negative B over 2a That's where the axis of symmetry is So let's say if you have a graph that looks like this. This equation will get this x coordinate.
And this is useful because, let's say if you plot this point, and you choose this point, they're going to have the same y value. And these two points will have the same y value. So if you want five points, you really need to find three. And two of them will be the same as the other two.
So a is 1. this is a and B is 2 so let's use the equation X is equal to negative B which is negative 2 over 2 times a or 2 times 1 so the vertex is that X equals negative 1 now we're going pick five points but our center point is going to be the vertex negative one we're going to pick two points to the right of that point and two points to the left two points to the left of negative one is negative two and negative three and two points to the right of it is zero and one So now let's plug in these numbers in. Let's start with negative 1. If we plug in negative 1 for x, we get negative 1 squared plus 2 times negative 1 plus 3. So that's 1 minus 2 plus 3, which is... 2 if we plug in let's say 0 for X it's just going to be 0 squared plus 2 times 0 plus 3 which is just 3 so 0 and negative 2 are going to have this same y value because they're equally distant from the vertex let's plug in 1 it's easy to plug in 1 then negative 3 so 1 squared plus 2 times 1 plus 3 that's going to be 6 so negative 3 also is going to have the same y value 6 all right so let's make our graph The majority of the graph is at the top. So, 1, 2, 3, 4, 5, 6. And most of our x values are on the left side.
So, 1, 2, 3, 4. 1, 2, 3, 4. So, the first point is at 0, negative 1, 2, which is here. That's the vertex. Next we have the point and.
And then we have, which is up here somewhere, and. So now we can graph it, and it's going to look like this. I can't believe I missed this point.
so that's a rough sketch the axis of symmetry is this line right here it's simply x equals negative one is the x coordinate of the vertex that's the AOS axis of symmetry if you need to find the minimum value the minimum value is the y coordinate of the vertex is right there that's where the minimum is located let's try another example like this for the sake of practice So let's say if we have y is equal to 2x squared minus 4x plus 1. So let's find the vertex first. It's negative b, which is negative 4, over 2a. a is 2. So this is a, this is b. So this is equal to 1. So let's make our table. And we need two points to the left of 1. That's 0 and negative 1. And two points to the right, 2 and 3. So let's begin by plugging 1 for x.
So 2 times 1 squared minus 4 times 1 plus 1. So this is 2 minus 4 plus 1. And that's equal to negative 1. Now it's easy if we plug in 0 instead of 2 So 2 times 0 squared minus 4 times 0 plus 1 is 1 The next point we're going to plug in is negative 1 this is also going to be 1 by the way So this is negative 1 squared is 1 times 2. Negative 4 times negative 1 is 4. So we're going to get 7 for this point, which means this one is also going to be 7. Okay, so now let's make our graph. So it's going to go up to at least 7. And most of our x values will be on the right side. So our vertex is at, which is over here.
Our next point is at, and we have another one at. And then, which is all the way here, and, which is right where I put the 7. So then we can graph it. And here is the axis of symmetry.
It's x equals the x coordinate of the vertex. So that's how you can plot quadratic equations using a table. But now let's say if we don't want to make a table. Let's say if we want to graph it a simpler way.
So let's say you have y is equal to x squared minus 2x plus... 3 Find the vertex as usual. It's negative B over 2a so it's negative Negative 2 over 2 times 1 so therefore X is 1 Now you want to find the y coordinate at that point so if you plug in 1 for watt for X you get 1 squared Minus 2 times 1 plus 3 that's 4 minus 2 which is 2 so we got the point 1 comma 2 So our graph is initially right here.
So here's what you can do. Let me show you a pattern. Let's say if you plug in 1 for x, you're going to get 1 for y, right? If you plug in 2 for x, you're going to get 4 for y.
if you plug in 3 for X you're gonna get 9 for Y so once you have your vertex and this is 1x squared by the way because if was 2x squared you have to double everything to find the next point after you have the vertex move once to the right and then up one and do the same thing for the left side once to the left up one because 1 squared equals 1. Now, 2 squared is 4. So starting from your vertex, as you move 2 to the right, go up 4. 1, 2, 3, 4. Same thing here. 2 to the right, up 4. And that's a quick way you can plot this equation without needing a data table. so now we're going to focus on graphing these equations but in vertex form so let's talk about how to convert a quadratic equation from standard form to vertex form now there's something called completing the square so here's what you need to do you have x squared plus 6x we're going to focus on that number six take half of that number and then square it half six history and we're going to square it we still have the minus 3 now notice that we added 9 to the right side so we change the equation to balance that we need to 9 to the left side or subtract 9 from the right side since I don't want to change the left side I'm just going to put minus 9 on the right side so if you add 9 and subtract 9 you haven't changed the value of the equation imagine if you have a bank account let's see have a thousand bucks in your bank account if you add a hundred and take away a hundred it's still the same so the two equations are equal to each other but this can be factored if you factor it here's a shortcut method it's going to be what you see here X and whatever this sign is plus whatever this number is before you square it 3 squared it's always gonna work out that way and then these two numbers you can combine so now you have an inverse x form which is x minus h squared plus k the vertex is h comma k So, in this form, you can easily see what the vertex is. So, in our example, notice that h and negative h, they have the opposite sign.
So, you've got to switch it from 3 to negative 3. This is h. And k, the sign is the same. It's both positive. So, whatever this number is, that's k.
You don't change it. So it's negative 3, negative 12. I'm going to go by twos. 2, 4, 6, 8, 10, 12. So that's negative 12. 2, 4, 6, 8, and so forth.
So we have the point negative 3, negative 12, which is somewhere in this region. And since 1 squared equals 1, as you move 1 to the right, go up 1. So that's going to be somewhere over here at negative 2 slash negative 11. 1 to the right, up 1. And then as you go 2 to the right, you need to go up 4. So from negative 12 is going to be at negative 8 and then 2 to the right up 4 Now if we go 3 to the right 3 squared is 9 we need to go up 9 so That will take us to negative 3. And 3 to the right here, which will be at negative 6, with the y value will also be at negative 3. If we go 4 to the right, 4 squared is 16. So we'll be at, the vertex is at negative 3. So 4 to the right, x will be 1. Negative 3 plus 4 is 1. And negative 12 plus 16 is 4. So we'll be at 1, 4, which is somewhere over here. and if we go forth to the left negative 3 minus 4 that's negative 7 and the Y value should still be the same it should be around 4 as well and now we can graph it so that's a rough sketch all my sketches are rough so that's what you can do if you have it in vertex form let's try another example So let's say if you have y is equal to negative x minus 3 squared plus 4. Now, this negative sign tells you that the graph is going to open downward.
So our vertex, change negative 3 to positive 3, but don't change the 4. That's our vertex, 3, 4. So let's plot that point. So here's 3, 4, and we're going to need more points on the right side. Now, we know the graph is going down. So as we move 1 to the right, it's going to go down 1. one to the left down one to the right it's going to go down for from starting from the vertex so it's going to be at 0 on the x axis 2 to the left is also going to be down for now if we go 3 to the right from the vertex which will be over here we need to go down 9 so we should be at negative 5 which is over here and if we go 3 to the right will be at the y-axis and will be down 5 as well so this graph looks something like this that's how you graph it. Now what about this one?
Let's say there's a 2 in front of the parentheses. So how does it modify our process? So the vertex we know it's negative 1, you change the plus 1 to minus 1, comma negative 2. This is going to stay, k is going to be the same.
my line didn't have a straight. So, notice that it's positive 2. So that means the graph is going to open upward. So let's plot the vertex.
So it's at negative 1 comma negative 2, which is over here somewhere. Now, we won't be moving 1 to the right and up 1 because our parent function is y equals 2x squared. There's a 2 in front.
So if you plug in 1 for x, you're going to get 2 for y. If you plug in for x you're going to get a for y so as we move one to the right it's not going to go up one it's going to go up two because it's going to be times two so one to the right up two that should be over here somewhere one to the left up two starting from the vertex as we go to the right we need to go up 8 instead of up 4 2 squared is 4 but we got to multiply by 2 so if we go up 8 we should be at 6 from negative 2 and 2 to the left we need to go up 8 as well it's going to be at the same point and so that's how you graph it And the axis of symmetry, if you have to find it, it's x equals this value, negative 1. So here's the line of symmetry. It's symmetric about that line. And this graph has a minimum. Whenever you have a positive line, you have a minimum.
number here the graph always has a minimum and the minimum value is simply the y-coordinate of the vertex is negative 2 when it opens downward let's say if you have negative x squared then you have a maximum sometimes you may find you may get like word problems where they ask you what is the maximum value of this function just find the vertex and look for the y value So that's it for quadratic equations. Now let's move on to absolute value functions. The absolute value of x looks like this.
It's a v-shape. And it has a slope of 1. So let's say if we want to graph the absolute value of x minus 3 plus 4. First, we need to find the vertex. So, we need to move 3 to the right, because we have a minus 3 on the inside.
And up 4. Kind of like the last problem, the vertex is 3, 4. You change what's on the inside, but you don't change what's on the outside. Now, the slope is 1 for this problem. So, as you move 1 to the right, you need to find the vertex. to go up one.
One to the left, up one. It's symmetrical, just like the quadratic function. But notice that it's y equals x rather than y equals x squared. So when x is one, y is one. When x is two, y is two.
So the slope is going to be one. And then just keep moving one to the right and up one. And then just plot the graph. So that's how you can graph absolute value functions.
but now let's say if we have y is equal to negative 2x plus 1 minus 3 Now, the fact that it's negative on the outside means that it's going to open downward instead of upward. And the fact that we see a 2, the slope is now 2 instead of 1. So as we move 1 to the right, it's going to go down 2. And the slope is constant. It's going to always be 2. So first, let's plot the vertex, which is negative 1, negative 3. so it's over here somewhere negative 1 negative 3 and we said that's going to go downward so as we move one to the right it's going to go down to you let me use a different color and one to the left and down to you and one more time one to the right down to you one to the left down to you so it's gonna look something like this so that's how you can graph absolute value functions so let's move on to cubic functions let's say if you have y is equal to X cube the parent function for this graph looks something like this that's just a rough sketch so let's say if you want to graph this equation The center point, where it should be, the origin, it's at 1, 2. You change the negative 1, but not the 2. So the shift, it moves 1 to the right, and it moves up 2. What you see on the inside is the horizontal shift, which is 1 to the right. This is the vertical shift. It's up 2. So the origin is now at 1, 2. Now we can draw a rough sketch if we want.
We can do something like... If we want to, but let's actually get points and make this graph more accurate. So... Our parent function is y equals x cubed. 1 to the 3rd is 1. 2 to the 3rd is 8. So to get the first two points, as you move 1 to the right, go up 1. 1 to the left, go down 1. Now if you move 2 to the right...
you need to go up 8, and we really don't have the space for that, but it's probably going to be somewhere over here. And if we move 2 to the left, we need to go down 8. Now this we can plot, because that's somewhere, that's going to be at negative 6. So down 8 will be all the way over here. So now we can get a more accurate sketch. So it's going to look something like that. So let's try one more example with cubic functions.
So let's say if you have y equals, instead of putting negative 2, I'm going to put negative 1 half because I don't want to go up 16. Let's say if we have this, and it's minus 1. We know the center is going to be at negative 3, negative 1. So it's going to shift 3 units to the left and down 1. So it's at negative 3, negative 1, which is in this area. And let's say if we move one unit to the right. Oh, by the way, notice that it's negative.
Positive x cubed looks like this. negative x cubed looks like that. So we have a decrease in function rather than an increase in function. So as we move one to the right, one, this is supposed to be cubed, by the way, one to the third, is 1 but we got a half so we should move down a half so one to the right it's going to go down a half because of this number one to the left is going to go up a half now two to the third is eight but times a half is four so as we move two to the right it's going to go down four so from negative one so negative five and as we move two to the left this point should be actually here let me fix that one to the left should have been up a half two to the right that's down four so it's over here and negative 5 and to the left we need to go up 5 I mean up for so 1234 should be over here somewhere so now let's see if we can make our graph So we have to be, the function has to be decreased instead of increasing, only because of the negative sign. So that's a rough sketch for that graph.
So now let's go over radical functions. Let's say if we have this function, the square root of x. The parent function for that looks like this.
And if you have, let's say, negative square root x. The negative in front causes it to flip over the x-axis, or it's a reflection across the x-axis, so it looks like this. And let's say if you have a negative on the inside. It reflects across the y-axis, so it looks like that.
And if you have a negative both on the inside and on the outside It reflects across the origin, so it looks like this The way I like to see this let's say if um Here X is positive and Y is positive positive so X and Y are positive in quadrant one so it goes towards quadrant one here X is positive but Y is negative so that's quadrant four X is positive as you go to the right but Y is negative as you go down here X is negative Y is positive that's quadrant two and when X and Y are both negative it's quadrant three that's a an easy technique you can use that to see where the graph is going to go for a radical X but now let's apply it in a problem so let's start with a straightforward one let's say if we have y is equal to the square root of x plus 3 minus 2 so the origin it's shifted 3 to the left and down to so it's going to start over here now the parent function is the square root of x The square root of 1 is 1, and the square root of 4 is 2, and the square root of 9 is 3. So as we move 1 to the right, go up 1. As you move 4 to the right, go up 2. So that's going to be somewhere over here. And as you move 9 to the right from your original point, go up 3. So right now we're at negative 3. If we move 9 to the right, we should be at positive 6. And going up 3, we need to be at y equals 1. So that's our graph. It starts from here.
Now one thing I should have mentioned is domain and range. What is the domain for this function? And what's the range?
The domain is the x values. The range is the y values. If you want to find a domain, analyze it from left to right.
So, it starts at the lowest x values at negative 3. There's nothing before that. And the highest x value, it goes to infinity. So your domain is from negative 3 to infinity, and it includes negative 3. The range is the y values.
the lowest y value is that negative 2 which is that number and this arrow it keeps going up but it goes up slowly and slowly but it goes up to infinity so the range is from negative 2 to infinity Now just to review, for linear equations like let's say y is equal to 2x plus 3, I'm just going to draw a rough sketch. So it starts at 3 with a slope of 2, let's say it looks like this. The domain for a linear function is just negative infinity to infinity.
The range is the same thing. Now the next one that we considered was a quadratic function. Let's say if you have 3 plus x squared.
The graph is going to start at 3, and it's going to open upward. It's been shifted 3 units up. So, the domain for any quadratic function, x can be anything. It's always going to be negative infinity to infinity. However, the range will change.
Notice that the lowest y value is 3, but it goes up to infinity. So the range is from 3 to infinity. Now if you have a downward parabola, let's say like y is equal to 5 minus x squared.
Your graph is going to start at 5. 5 and it's going to open downward it's been shifted up 5 units so your domain for any quadratic function where you have x squared as your your highest degree it's always going to be negative infinity to infinity the range though it's from let's see your lowest y value is negative infinity because these arrows will go all the way down your highest y values at 5 so from low to high negative infinity to 5 it stops at 5 So for your domain, you're looking at it from left to right. From the range, you're looking at it from low to high, or bottom to top. Now the next thing that we considered was an absolute value function. So let's say if you have 2x minus 3 plus 1. So the graph shifts 3 units to the right, up 1, it starts here, and the slope is 2. So I'm just going to draw a rough sketch.
Your domain for any absolute value functions is negative infinity to infinity. Unless you have like a piecewise function. Your range, notice that your lowest y value is 1, but it goes up to infinity.
So it's from 1. one to infinity and let's say if it opens downward let's see you have three minus absolute value of X plus 2 so it shifts two units to the left and it's shifted up three units so starts here but there's a negative in front of the absolute value so it goes like this it goes downward with a slope of 1 so your domain is everything for this function as well but the range the highest value is 3 the lowest value is negative infinity starts from low to high negative infinity and stop at three now for cubic functions say if we have X minus 1 cube plus 2 as you move one to the right it's going to be shifted up to and there's no number in front of in front of here so it's just a 1 and because it's positive it's going to be increasing rather than decreasing so It's going to look something like this. For this type of function, your domain is also everything. There's no restrictions on x. Your range is also everything as well.
It goes down to negative infinity and up to infinity. So for any cubic function, domain and range is just R-row numbers. So now let's go back to radical functions. Let's look at our next example. Let's say if we have the square root of 5 minus 2 minus x.
so the first thing we want to plot is what the first point so if you set the inside equal to 0 2 minus x is equal to 0 you're going to get x is equal to 2 so it shifts to units to the right and it shifts up 5 units So it starts somewhere here. The question is, what direction is it going to go? Is it going to go towards quadrant 1, towards quadrant 2, towards quadrant 3, or towards quadrant 4?
Because knowing the direction is important. we need that if we don't want to make a table. So notice that we have a negative in front of the radical and a negative in front of X.
Do you see that? So X is negative and Y is negative. X and Y are negative in what quadrant? has to be quadrant 3 so it's going to be going in this general direction so now we don't have a number in front of the radical so we can use our general transformation the square root of 1 is 1 The square root of 4 is 2, and the square root of 9 is 3. So as we go one unit to the left, it's going to go down one. And as we, this is x, this is y.
So as we travel four units to the left, it's going to go down two. So it's going to be over here. And as we travel nine units to the left, starting from this point, It's going to go down 3, so it's going to be over here. Now for this, if you prefer, you can always make a table. So I'm going to do one example by making a table so you can see which points to choose.
And there is the graph right there. So let's say if you have y is equal to the square root of x plus 1 minus 3. And you want to make a table. The first point you want to plug in is negative 1, and it's going to be negative 3. That's the, because that's...
shift shift one to the left and down three now from that point let me just put this number here You want to go 1 to the right, so let's plug in 0. If you plug in 0 for x, the square root of 1 is 1, and 1 minus 3 is negative 2. Now your next point, you don't want to plug in 1, 2, 3, and 4. You want to plug in numbers that are perfect squares, because you can square root it. The reason why I plug in 0 is because I can square root 1. The next number I'm going to choose is 3, because 3 plus 1 is 4, and the square root of 4 is 2. 2 minus 3 is negative 1. now the next point I'm going to plug in is 8 because 8 plus 1 is 9 the square root of 9 is 3 and 3 minus 3 is 0 if I want another point I'm going to plug in 15 because 15 plus 1 is a perfect square 16 the square root of 16 is 4 4 minus 3 is 1 so you want to plug in numbers that are perfect squares where the radical is simplified to an integer and that's going to make your life easier and basically the way I was plotting points for these other functions is based on this pattern of numbers so let me show you so we know it shifts one to the left down three so this point is this one negative 1 negative 3 and for radical X Radical 1 is 1. So as we move 1 to the right, it's going to go up 1. Notice that we have the point 0, negative 2. And the square root of 4 is 2. So starting from this point, if we move 4 to the right, which will put us at 3, it's going to go up 2. So notice we have this point, 3, negative 1. And then the square root of 9 is 3. So as we go 3 to the right, So 1, 2, 3, 4, 5, 6, 7, 8, 9. We're going to go up 3, so it's going to be over here, which is 8, 0. So notice this technique gives you the points without actually making a table. And if you can master this technique, you can graph equations at a very fast rate.
So our graph looks like this. Now, keep in mind, let's say if we put a 2 in front, everything is going to double. So, as you go 1 to the right, you're going to go up 2 instead of up 1. As you go 4 to the right, instead of up 2, you need to go up 4, which is somewhere over here. And as you go 9 to the right, instead of going up 3, you need to go up 6. which will be somewhere over here so your graph will look something like this if it was y equals 2 square root x plus 1 minus 3 So the amount that you increase in your y values, just double it.
That's it. If there's a 2 in front. If there's a 3, just triple it. So like if we use this pattern here.
Let's say if there was a 2. It would be 2 square root of 1, which is 2. If we plug in a 4, it's going to be 4. If we plug in 9, it's going to be 6. Square root of 9 is 3 times 2. to a six so these numbers the amount that we increase the y values was simply double Okay, so now let's move on to a different type of radical. A radical with an odd root instead of an even root. The square root of x, what we considered was this one.
There's an invisible 2 that you don't see. So if it's even, it looks like this. But when it's odd, it looks like this instead. so negative cube root of x is going to look like this by the way keep in mind this is the same as cube root of negative x so even if you reflected across the x-axis or across the y-axis it's going to be the same because this graph is symmetrical about the origin so there's only two ways you can graph cube root of x this way if it's positive or this way if there's at least one negative if there's two negatives they can cancel and it's going to be the same as this one if you have a negative on the inside and on the outside but now that we know what the parent function looks like let's Let's try some problems.
So let's say if we want to graph the cube root of x plus 1 plus 2. So first you want to find out where the graph begins. So it shifts 1 to the left and up 2, so it's over here. Now let's look at our pattern of numbers that we're going to use to plot this equation.
So the cube root of 1 is 1. So as you move 1 to the right, it's going to go up 1. cube root of 8 is 2, so as we move 8 to the right, it's going to go up 2. So, 1 to the right, up 1, 1 to the left, down 1. Because we know the general shape is like this. It's increasing. Now as we go 8 to the right, so starting from negative 1, we're going to stop at 7. It's going to be up 2, so it's going to be over here somewhere. And as we go 8 to the left, It's going to take us to negative 9. So it's going to go down 2, so it's going to be over here somewhere.
And so now we can just plot it. So it looks something like that. So that's how you can graph cube root functions.
Now let's say if you have 1 over x. The parent function for 1 over x looks like this. And if you have negative 1 over x, it looks like this.
It reflects over the y or the x-axis. It doesn't matter. It'll be the same.
So let's try some problems. Let's say if we wish to graph 1 over x minus 2. So we have the general shape, but you want to plot the vertical asymptote. The vertical asymptote is when the denominator is equal to 0. So if you set the bottom equal to 0, you get x equals 2. That's the VA, the vertical asymptote. Now there's also a horizontal asymptote, which is y equals 0, which is over here. Whenever the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote is at y equals 0. Now if you want to get points for this graph, there's really only one point you need.
But let's make a table. I would plug in 3 for x because 3 minus 2 is 1, and 1 over 1 is 1. So at 3, it should be at 1. The next point I would plug in would be something to the left of 2. Let's say like... If you plug in 1, you're going to get negative 1. So it should be somewhere over here.
So then our graph starts from the asymptote, and then it connects to this point, and then it follows the horizontal asymptote. so it looks something like that. You can add more points if you want.
And this graph looks like that. So that's how to graph 1 over x minus 2. So let's say if we have y is equal to 1 over x plus 3 plus 4. The vertical asymptote is at x equals negative 3, based on this term here, if you set it equal to 0. Now the horizontal asymptote for this fraction is 0 because it's bottom heavy. The degree of the denominator is higher than that of the numerator.
But it's 0 plus this number, 4. So it's actually y equals 4. It's been shifted up 4 units. So let's make our graph. So the vertical asymptote is at negative 3. And the horizontal asymptote is at 4. And let's plug in some points.
So I'm going to plug in one point to the right of negative 3, so that's going to be negative 2, and one point to the left of negative 3, that's negative 4. If I plug in negative 2, negative 2 plus 3 is 1, 1 over 1 is 1. If I plug in negative 4, negative 4 plus 3 is negative 1, and so I got those points. So negative 4, negative 1, it's around here. And negative You know what, I forgot to add 4 to each number. So the y value should be 3, and the y value here should be 5. Can't forget this 4. But I plotted at the right point for some reason.
At negative 2, it should be at 5, so somewhere over here. Now we can make our graph. So this one's going to look something like this, and this graph is going to look something like that. As you can see, it's been shifted left 3, up 4. Now let's say if we have a negative value. Let's say y is equal to negative 1 over x minus 1 plus 2. So the vertical asymptote, if you set x minus 1 equal to 0, x is equal to 1. The horizontal asymptote is going to be y equals 2, based on this number.
So here's the horizontal asymptote, and the vertical asymptote is that one. Now notice that we have a negative sign, so instead of the graph looking like this, it's going to be like this instead. So let's plug in points. Let's make a... make a table so I'm going to plug in 2 for X and 0 for X 2 minus 1 is 1 but times negative 1 divided by 1 is negative 1 plus 2 so that's 1 if we plug plug in 0, it's going to be negative 1 over negative 1 plus 2, and we're going to get 3. If we plug in 0 for x.
So we have the point, which is over here, and, which is here. So this graph is going to look like this, and this one is going to look like that. So now you know how to graph parent functions, 1 over x functions. Now let's say if you have 1 over x squared. It's very similar to 1 over x, but both curves are above the x-axis.
If you have negative 1 over x squared, it's going to be below. the x-axis. It's going to look like this. So this graph is symmetrical about the y-axis. The other one, 1 over x, was symmetrical around the origin.
So let's try one example. Let's say if we have negative of 1 over x minus 2 squared plus 3. The vertical asymptote, you set the bottom equal to 0, you're going to get x equals 2. The horizontal asymptote, it's simply y equals 3. So for these types of rational functions, always plot the asymptotes first. So let's make a table.
I'm going to plug in 1 for x and 3. 1 unit to the left and 1 unit to the right. So if I plug in 3, 3 minus 2 is 1, 1 squared is 1. Negative 1 over 1 is negative 1 plus 3, that's 2. If I plug in minus 2 is negative 1 but if you square it becomes positive 1 negative 1 over 1 is negative 1 plus 3 is 2 notice that they have the same y value which means they're symmetric around about the the y axis but in this case it's going to be symmetric about the vertical asymptote because it's been shifted to unicentivite. So if we plot 1, 2, that's over here. I mean, that's 3 common 2, not 1, 2. 1, 2 is over here.
But we can see the symmetry around the vertical asymptote. So it looks like that. So let's go over some different types of rational functions. let's say if you have x squared plus 5x plus 6 divided by x squared plus x minus 2 now this function might seem complicated but it's not really we've pretty much covered this material what you should do is you should factor everything so to factor x squared plus 5x plus 6 Think of two numbers that multiply to 6 but add to the middle term 5. 2 times 3 is 6, 2 plus 3 is 5, so it's going to be x plus 2 times x plus 3. To factor this one, look for two numbers that multiply to negative 2 but add to 1. So that's positive 2 and negative 1. 2 plus negative 1 is 1. So it's x plus 2, x minus 1. Notice that the x plus 2's cancel. So, if you set x plus 2 equal to 0, x equals negative 2 is a whole.
The vertical asymptote is based on this one, because the x minus 1 couldn't be cancelled. So, x equals 1 is the vertical asymptote. Now, you might be wondering, what is the horizontal asymptote?
Now, notice that the degree of the denominator is 2, and the degree of the numerator is 2 as well. When it was bottom-heavy, when the degree of the denominator was higher than that of the numerator, like 1 over x, or 1 over x squared the horizontal asymptote was y equals 0 plus if there was a constant here if it was plus 3 then add 3 but we don't have that 3 there so whenever the degree is the same divide the coefficients 1 divided by 1 is 1 so the horizontal asymptote is y equals 1 for this problem so now let's graph it So first let's plot the vertical asymptote at 1. Next let's plot the horizontal asymptote at 1. So now at this point, we know the graph is going to look like 1 over x. It's going to be very similar to it.
So we just need a few points. One point I would plug in is the whole negative 2, because that's important. If we plug in negative 2, we just need to plug it into the survival equation.
So it's negative 2 plus 3, negative 2 minus 1. negative 2 plus 3 is 1 and negative 2 minus 3 is negative 3 so we get the point negative 1 3rd I would also plug in one number to the right of the vertical asymptote as we did before like 2 let's see if I could fit it here 2 plus 3 is 5 and 2 minus 1 is 1 so we just get 5 and plug in one number to the left of it so let's say 0 0 plus 3 is 3 and 0 minus 1 is negative 1 so we get negative 3 so negative 2 comma negative 1 third somewhere over here and 2 comma 5 is up here somewhere and 0 negative 3 is down here so we can see how this graph It starts from one asymptote, it connects to these points, and it follows the other one. However, we did say this is a hole. So this should be an open circle. I know I drew a closed circle, but ignore the yellow point.
It should be a hole there. now for the other one we know what the general shape is going to be it's going to look something like this you can add more points if you want but that's a rough sketch of the graph so let's try another function let's say if we have y equals x squared plus let's make that a 8x squared divided by 4 minus 2x squared plus 1 so let's take out a 2 from the bottom notice that's going to be 2 minus x squared and we're going to set that part equal to 0 and we're going to solve for x so x is equal to plus or minus radical 2. So we have two vertical asymptotes in this problem. Now to find the horizontal asymptote, notice that the degree of the numerator, which is x squared, is the same as that of the denominator.
So what we're going to do is divide the coefficients, 8 divided by negative 2. So the horizontal asymptote is at negative 4, but we have a constant in front, so shift it up one. So negative 4 plus 1, so it's at negative 3. So now we have the vertical asymptote and the horizontal asymptote. The square root of 2 is about 1.4, so the vertical asymptote is somewhere between 1 and 2. And we have another one at negative 1.4.
And the horizontal asymptote is at negative 3. Now we know how the graph is going to look like on the right side and the left side. It can be like this. Or it can be like that.
It can be like this. Or it can be like this. In the middle, a lot of things could happen.
You can get something like this. You may get a curve that looks like this. Or sometimes it can even cross the horizontal asymptote. So for the middle, we've got a plug-in point to find out what's going to happen.
So, let's make our table. For this one, I would definitely use a table. Let's plug in one point to the left of radical 2. So, let's plug in positive 2 for x. So, that's going to be 6, well, 2 squared is 4, 8 times 4 is 32. And, let's see, on the bottom we have 4 minus 2 times 2 squared. So, that's like 4 minus 8, which is negative 4. and so we should get negative 8 so 2 comma negative 8 so it's like all the way over here all we need is one point in this region so we know the graph looks something like this in this area so let's plug in another point let's say like negative 2 notice that it's squared so regardless if you plug in 2 or negative 2 the y-value is going to be the same so because it's both squared it's going to be symmetric or symmetrical so a negative 2 is going to be like negative 8 again so this is going to be the same so the fact that it's symmetrical means that we're probably going to have something like this or something like this so let's plug in 0 first if we plug in 0 for X everything is 0 so it's over here let's say if we plug in 1 and 1 and negative 1 is going to be the same if we plug in 1 it's going to be 8 over 4 minus 2 so that's 8 over 2 which is 4 So that's like over here somewhere.
So we can see what's happening. This graph is going to look something like this at the center. So between two vertical asymptotes, you can't be certain what's going to happen, but if you see that both of these... terms are even, like it has an even exponent, you know that there's going to be some sort of symmetry. It can be like this, or it can be like that.
If it was odd, it could be like this, at the center, between the two vertical asymptotes. could be like that. You just gotta plug in points to find out for sure.
So that's it for that particular function. Now there's one more thing we gotta mention, and that's rational functions involving a slant or an oblique asymptote. So notice that the degree of the numerator, which is 2, is greater than that of the denominator, which is 1. When the numerator is 1 degree higher than the denominator, you're going to have a slant asymptote. now we should be first as we should factor everything before you begin I'm if we multiply these two numbers two times five is 10 and two numbers that multiply the positive 10 but add to negative 3 doesn't happen it doesn't work because 2 times 5 will add up to 7 negative 2 and negative 5 will add up to negative 7 so we can't factor the numerator if we could we should we know the vertical asymptote is going to be x equals 2 Now whenever the numerator is higher than the denominator, there's no horizontal asymptotes. But we do have a slant.
So we need to do long division to find a slant asymptote. But we don't have to complete the entire long division process. There's a certain point where we can stop, where it would suffice and we can have the answer. 2x squared divided by x is 2x.
After you divide, multiply. 2x times x is 2x squared. 2x times negative 2 is negative 4x. So let's subtract. So here we get 0, negative 3x minus negative 4x, which is like plus 4x.
You get x. And bring down to 5. x divided by x is 1. Once you get this constant here, that 1, you can stop. So the slant asymptote is 2x plus 1. So now we can graph it.
So at 2 we have the vertical asymptote. And to plot the slant asymptote, notice that we have a linear equation in slope intercept form, which we covered at the beginning of this video. We're going to plot the y-intercept, which is 1. And the slope is 2, because there's a 2 in front of x. Excuse me. So that's the move once the right go up 2. And that's the move once the right go up 2 as well.
And then we're going to plot it with... a dash line. Now these two asymptotes separate the graph into four regions. So we can have a graph that looks like this, it can look like that, it can be over here.
or it could be here. It's going to be in two out of those four regions. And the only way to find out which region it's going to be is to plug in points. We only really need to plug in two points.
One point to the right of the vertical asymptote and one point to the left. So let's say if we plug in 3 for x. 2 times 3 squared.
3 squared is 9 times 2 is 18. Negative 3 times 3 is negative 9 plus 5. So that's 9 plus 5. That's 14 on top. 3 minus 2 is 1. So we get 14. So at 3, it's clearly above the... It's way above the slant asymptote.
It's all the way at 14, which our graph doesn't go that high. So we're going to say it's in this region somewhere. We don't have enough space to plot it. Usually, if it's in that region, then chances are it's going to be in this region.
But let's find out for sure. So let's plug in a number less than 2. Let's plug in 1. So 2 times 1 squared is 2, minus 3 times 1, which is 3, plus 5, over 1 minus 2. So that's negative 1 plus 5. That's 4 over negative 1, so it's negative 4. So at 1, it's at negative 4, so it is in this region. Let's go ahead and plug in 0. 0 is an easy number to plug in. It's going to be 5 over negative 2, which is like negative 2.5, just to make this graph a little bit more accurate. So our graph is going to look something like this.
Now the more points you add, the more accurate your sketch will be, but it's up to you how many points you wish to add. But usually on a test, you might get a multiple choice question, and all you need is one point to the left and one point to the right of the vertical asymptote, and you could eliminate the wrong answer. answers to get the right answer or you could just use a graph calculator so now we're going to focus on graph in exponential equations so let's say if we have y is equal to 2x plus 3 minus 1 Now, what you want to do is set the exponent equal to two numbers, 0 and 1. So, if you solve for x, you get negative 3 and negative 2. These are the points that you should plug in into your x and y table, which I would recommend for this type of problem.
So if you plug in negative 3 for x, negative 3 plus 3 is 0. 2 raised to 0 is 1. Anything raised to 0 powers 1. And then 1 minus 1 is 0. If you plug in negative 2 for x, negative 2 plus 3 is 1. And 2 to the 1 is 2. 2 minus 1 is 1. Now this number that you see here, this is the horizontal asymptote for an exponential function. So we're going to plot that. So we have it at negative 1. And there's no vertical asymptotes for exponential functions. Your next thing, after you plot the horizontal asymptote, plot your two points. Negative 3, 0. And negative 2, 1. So the graph, it starts from the horizontal asymptote, and it follows those two points.
And that's how you can graph it. You can add more points if you want, but I think that sketch is good enough. So let's try another one. So let's say if we have y is equal to 2 minus 3 raised to the x plus 1. So set the exponent x plus 1 equal to 0 and 1. And solve for x. So x is equal to negative 1 and x is equal to 0. And then make your table.
So we're going to plug in negative 1 and 0 for x. If we plug in negative 1, negative 1 plus 1 is 0. And 3 to the 0. is 1, 2 minus 1 is 1. If we plug in 0, 0 plus 1 is 1, 3 to the 1 is 3, so we have 2 minus 3, which is negative 1. This number is the horizontal asymptote. It's the constant that doesn't have an x.
So the horizontal asymptote this time is at 2. and if we plot our points negative 1 comma 1 and 0 negative 1 so start from the horizontal asymptote and connect the two points so that's how you can graph that particular exponential equation so the last thing we're going to look at is graphing logarithmic equations so let's say if you have log base 2 X plus 1 minus 3 So, exponential functions are like the opposite of logarithmic functions. Exponential functions have horizontal asymptotes, logarithmic functions have vertical asymptotes. Now, what you want to do is you want to set the inside part of the log function equal to three things rather than two things.
Set it equal to zero. which will give you the vertical asymptote always set it equal to 1 and set it equal to whatever this number is the base so the first one we get X is equal to negative 1 that's the vertical asymptote the next point we get 0 and here we get one so let's make our table with these two points 0 and 1 if you plug in 0 for X you get 0 plus 1 log of 1 regardless of what the base is is always 0 So you have 0 minus 3 is negative 3 if we plug in 1 1 plus 1 is 2 So we have log base 2 of 2 that 2 is canceling to give you 1 and 1 minus 3 is negative 2 so that's why I chose those numbers because You get a nice whole number integers to deal with which makes the easier to graph So the first thing you want to do is plot the vertical asymptote, which is that negative 1. and then plug in your points so we have our first point at 0 negative 3 which is over here and then 1 negative 2 so start from the vertical asymptote and then follow the two points and that's how you graph it let's try our final example so let's say y is equal to 2 minus log base 3x minus 1 so set the inside equal to 0 1 and the base in this case which is 3 so the vertical asymptote is X is equal to 1 and our other points are 2 and 4 so let's make our table So if we plug in 2 for x, it's 2 minus 1, which is 1 log of 1 is 0, so 2 minus 0 is 2. If we plug in 4, 4 minus 1 is 3, log base 3 of 3 is 1, and so 2 minus 1 is 1. So our vertical asymptote is at 1. And we have the point, which is over here. And the next one is.
So the graph has to start this way if it's going to connect to those two points. And that's how you plot it. So that's it for this video. Now you know how to draft almost any equation. There's other equations too, but for the most part, we covered the most common equations that you'll see throughout your algebra, trigonometry, pre-calculus, and calculus course.
So if you see any new equations that you have to graph, you can apply the principles that you've learned in this video to those equations as well. So that's it, and thanks for watching this video, and have a great day.