Today we're going to talk about graphing linear equations. So we're going to do three different examples. First graphing a line given an equation written in slope intercept form, then graphing a line given equation written in slope point form, and finally, graphing a line given an equation written in standard form.
So let's start by looking at how to graph an equation that's written in slope intercept form. So I have the equation y is equal to 2x minus 4. And thinking about the form of slope intercept, I know that my slope is going to be 2, and my y-intercept is going to be negative 4.
So I'm going to start first by plotting the points for my y-intercept. So I'm going to go to my y-axis, and down to negative 4, and place a point. Then I'm going to use my slope and calculate it by thinking about the slope as rise over run to get another point on this line.
So if my slope is 2, I know that I can rewrite that as a fraction as 2 over 1. So my rise of this line is going to be 2. And the run of this line is going to be 1. So from, again, my y-intercept, I'm going to go up 2 and over to the right 1.
I could do this again by going up 2 and over 1. And I can continue doing this as many times as I needed to, or wanted to to create my line. But you only need two points to connect to be able to draw a line of an equation.
So now that I have my point, I'm just going to go ahead and connect them. So I can see that the graph of the equation 2x minus 4 is a line that looks like this.
So let's look at how we can graph an equation that is written in slope point form. So I have this equation in slope point form, and I know that I can determine a point that the line passes through, and the slope of that line. So I'm going to start by determining the point that the line passes through.
So I know that my two values for my point are going to be here and here. And I need to be careful because in the general form, or the formula for an equation written in slope point form, I have x minus my x1 value. So because this is a plus 1, that would be the same as minus a negative 1.
So the x value of my point is actually going to be negative 1. And then here, I know that the point-- the y value of that point is going to be positive 4. Again, because in the general form, or the formula for an equation written in slope point form, I have y minus my y1 value, or my y value. So this is just going to be 4.
So the point that this line passes through is going to be negative 1, positive 4. So I'm going to start by plotting that point-- negative 1, positive 4. And now that I have my point, I can start from that point and calculate-- or use the slope as rise over run to determine another point on that line.
So my slope is negative 3 over 2. And again, if I think of that as rise over run, I know that from this point I'm going to go down 3, and over 2 to get another point. So I'm going to go down 3 and over 2. And I find that another point on that line is at 1,1.
So I can continue using the slope of negative 3 over 2-- to 3 over 2-- to find more points on my line. But again, you only need to use two points to be able to graph the line.
So now that I have at least two points, I have more than two points, I can go ahead and connect them to make my line. So once I do that, I have a line that looks like this.
So finally let's see how we can graph the equation of a line that's written in standard form. So I have this equation negative 2x plus y is equal to 5. And I can see that this is written in standard form.
And I know that in standard form, we can't easily, just by looking at the equation, determined the slope, intercepts, or points that lie on that line. So I'm going to go ahead and use the equation in standard form to find the x and y-intercepts. And then I can connect those two points to get the graph of my line.
So I'm going to start by looking for my x-intercept. And I know that my x-intercept is the point where it crosses the x-axis-- the line crosses the x-axis. And this point is also when y is equal to 0. So I'm going to substitute 0 into my equation for y, and then solve for x.
So doing that, my equation becomes negative 2x plus 0 is equal to 5. That will give me negative 2x is equal to 5. And then, to isolate my x variable, I'm just going to divide by negative 2. This will cancel, and I get that x is equal to 5 over negative 2, which is just negative 2.5.
Similarly, to find my y-intercept I know that my y-intercept is where my line crosses the y-axis. And it's also where x is equal to 0. So now I'm going to substitute 0 in for my x variable in my equation, and solve for y.
So now I will have negative 2 times 0 plus y is equal to 5. Negative 2 times 0 is just 0. So I have y is equal to 5. So my y-intercept is 5.
So now that I know my two intercepts, I can go ahead and plot these two points on my graph. So x is equal to negative 2.5 will be about here. And y is equal to 5 will be here.
And because I only need two points to draw a graph of my line, I can go ahead and just connect these two points. So doing that, I can see that the graph of the equation negative 2x plus y equals 5 looks like this.
So let's go over our key points from today. As usual, make sure you get them in your notes so you can refer to them later. To graph the line, we only need to know the coordinates of two points on the line.
And points on a line can be determined given a starting point and the slope, which is again the rise over run. So I hope that these key points and examples helped you understand a little bit more about graphing linear equations. Keep using your notes and keep on practicing, and soon you'll be a pro. Thanks for watching.