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Graphing Quadratic Equations - Problem 1 7,266 views

Teacher/Instructor Alissa Fong
Alissa Fong

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

Here I have a parabola that I’m asked to graph. I know it’s going to be a parabola because I have a quadratic equation, meaning my highest exponent on x is 2, it’s an x² graph. So I could go through and make a table of values, but what I’m going to do instead is try finding the important points like the x and y intercepts and the vertex, maybe the Axis of Symmetry.

Let’s start with the y intercept. The way to find the y intercept always is to let x equal 0, so what I’m going to do is substitute 0 into my function and see what my y value is, or my f(x) value. If I put in 0 there, I’ll have 0 take away 0 plus 3, so that tells me my y intercept is at 0 for x, 3 for y. That’s one important point.

The next thing I’m going to look for is the x intercepts. Now when you’re looking for the x intercepts, you have lots of possible methods. You could do completing the square, you could use the quadratic equation, you could use taking square roots of both sides if there is no b term. I’m going to try factoring because I have a little more practice than you guys and I can tell that this guy is going to factor nicely.

So I’m going to set it up letting y equal to 0, go through and factor this and then using the zero products property, I’ll be able to find my x solutions. Two numbers that multiply to +3 and add up to -4 would be -3 and -1. Set each one of those quantities equal to 0 and I’ll have x equals 3 and x equals 1. That tells me that my two x-intercepts are going to be the points (3,0) and (1,0). I have my x intercepts, I have my y intercepts that’s great, because those are the important points.

I want to find out some more of those, so I’m going to look for the vertex. The way to find the vertex is two steps. The first thing for the vertex is to find the x coordinate. The way you find the x coordinate is by finding -b on top of 2a. My b value is -4, so I’m going to write –(-4) which is like +4, 2 times my a value is 2×1, so the x coordinate of my vertex is going to be 4 over 2 which reduces to 2.

To find my y value, I’m going to substitute in x equals 2 right here and see what my y output is. Y is going to be 2² take away 4 times 2 plus 3, so -4 plus 3, y is going to be -1. There is my vertex. That’s the point my parabola is going to change; the y values are going to change from either increasing to decreasing or from decreasing to increasing, which brings me to thinking about whether this parabola opens up or opens down. You can tell by looking at a, the leading coefficient. Here my a value is 1, that tells me a couple of things. It tells me the parabola is going to open up because a is positive. It also tells me it’s not going to be very wide or skinny, it’s just like regular parabola width and that will make more sense to you as you start graphing more and more parabolas.

Okay, so I have a few points that are pretty important, I’m going to put those guys on the graph. I may or may not choose to find more x values to put on my graph as well. First thing I’m going to put on there is the y intercept which is the point (0,3) there it is. Then I’m going to stick on there my x intercepts which are 3 and 1, on the x axis, then I’m also going to put on there my, wait is that 3? 1, 2, 3 there we go oops. I’m also going to put on there my vertex. My vertex is the point (2,-1) so I’ll go 2, -1, good.

So I’ve got 4 points on there I can kind of tell where my parabola is, but in order to have a really accurate graph, I’m going to use the Axis of Symmetry. Remember the Axis of Symmetry is this vertical line that passes through your vertex and it has the equation x equals –b over 2a. We already did –b over 2a, and we found out that was 2. So that line x equals 2 is my Axis of Symmetry and that helps me get some more points like for example this point right here, I know is reflected across the Axis of Symmetry and would show up there. That helps me have a better idea of what my parabola looks like.

So most teachers will let you graph your parabola using 5 points like this. Some teachers might as you to find more than 5 points in which case you’d have to substitute in x equals 4, or x equals -1 or whatever you want. You can kind of tell if I plugged in x equals -1 I get some value up here or up there for 4. It’s up to you and your teacher how precise you need to be, but using my y intercept, my x intercepts, my vertex and my Axis of Symmetry as well as what I knew about my a leading coefficient, it told me that parabola opened up and that it was kind of medium. It wasn’t skinny or wide.

All of that information helped me create this graph which I think looks pretty good. So there’s lots of step to this. You guys will get better and if you have a graphing calculator at home it’s good idea to check your work just make sure you did everything correctly.