MA, Stanford University
Teaching in the San Francisco Bay Area
Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts
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MA, Stanford University
Teaching in the San Francisco Bay Area
Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts
To graph a line using the standard form of a line, find the x and y-intercepts. To find the x-intercept, substitute in 0 for y and solve for x. The resulting value for x will be the x-intercept, of the point on the x-axis where the graph of the line will pass. To find the y-intercept, substitute in 0 for x and solve for y. The resulting value for y will be the y-value of y-intercept, or the point on the y-axis where the graph of the line will pass.
Here I'm asked to graph an equation that's given to me in standard form. A lot of people choose to graph problems like these by finding the x and y intercepts and that's absolutely a valid method. The way you would do that is substitute in 0 for x to find the y intercept. Substitute in 0 for y to find the x intercept and then connect those two points.
I want to show you another way using Slope-Intercept. It's my personal favourite method because for me it's the quickest. Even though this equation is not in Slope-Intercept form, I can get it into Slope-Intercept form without too much difficulty. Slope-Intercept form remember look like this where y is all by itself. So first thing I need to do is get rid of this -2x by adding 2x to both sides of the equal sign, so I'll have -4y equals +2x plus 8.
Then since I want y all by itself, I'm going to divide by -4 all the way across, so now I'll have y equals 2 over -4 reduces to -1/2 and then 8 over -4 reduces to -2. This is the same equation as that original one only written in a different form and I like this form because I can graph it using Slope-Intercept.
Here is how. My first dot will go at -2 on the y axis because that's the y intercept. From there I'm going to count the slope which is going to be down 1, right 2 because it's negative. I'll show you what I mean. Our first dot is going to go at -2 on the y axis, so here we go y axis -2. From there, I need to show a negative slope, so instead of going up 1 over 2, I'm going to go down 1 over 2. I'm going to draw a couple points in this direction so I can make sure my line is correct when I get the ruler up there.
You could also draw a negative slope form here by going up 1, but then to the left 2. This is the same slope. It's still a slope of -1/2. I have that constant ratio all the way across my line. All I have to do left is connect the little points and then stick arrows on the end. Connect them like so, make a sound effect if you want to, and then stick arrows on the end.
Okay that's an A+ graph that I did pretty easily. You could have used intercepts to solve this problem, but my personal preference, the way that I make the fewest errors, is by putting into y equals mx plus b form and counting the slope after I start at the y intercept.
Unit
Linear Equations and Their Graphs