# Graphing Quadratic Equations - Problem 3

I want to go to graph this parabola, but the first things I’m going to want to do is find some important points like the y intercept, the x intercepts, the vertex, the Axis of Symmetry blah, blah, blah.

Okay the y intercept is the easiest for me so I’m going to start there. Always the way to find the y intercept is to let x equal 0. So the y intercepts when I put in x equals 0 in this case, my y value is 5. That’s one of the points on my graph. Now I’m going to look for the x intercepts. I personally like factoring, so I’m going to try to factor this guy. I know I’m going to start with x and x and then I need two numbers that multiply to +5 and add up to -4. Multiply to +5, so it has to be either 5 and 1, or -5 and -1, but I’m not going to get things that add up to -4, shoot. I think this cannot be factored. I think there might be no x intercepts.

To be sure what I’m going to do is find the Discriminant. The discriminant, remember b minus 4ac tells you how many solutions there are. I think there are zero solutions because this cannot be factored. Let me try it out. If I do b² minus 4ac, I get -4² take away 4 times 1 times 5, and I have 16 take away 20. I have a negative discriminant. That tells me there are no x intercepts.

You might also have tried the quadratic formula. If you try to do the quadratic formula with this a, b and c values, you would have to do the square root of a negative number. That of course is not possible. It’s no real solutions, so that tells me there is no real x intercepts.

Okay so I’ve tried y intercepts, I tried x intercepts, let’s look for the vertex. The way to find the x value of the vertex is to do -b/2a, so in my case, I’m going to do +4 because I changed the sign on top of 2 times 1. My x value is 2. Once I found the x coordinate, my vertex, let’s find my y coordinate, y is going to be 2² which is 4 take away 4 times 2 plus 5, so I have 4 take away 8 plus 5, that’s -4 plus 5 is 1. Okay so here is my vertex (2,1).

This is going to be a little bit tricky I want to go ahead and start doing some graphing, but I only know two points. I know the y intercept, and I know the vertex, so I’m still kind of nervous about this graphing situation. Let me see what else I know before I start graphing and we'll see if we can come up with anything else to help.

I know this parabola is going to open up because my leading coefficient is positive. I also know that my parabola is going to be not too wide, not too skinny because my leading coefficient is 1. So let’s put some points on the graph, and try to find the Axis of Symmetry and use that and maybe also pick another x value to find the corresponding y value.

The first thing I’m going to put on my graph is the y intercept (0,5) 1, 2, 3, 4, 5, I’m also going to put on here my vertex (2,1) there it is. That’s all I know so far, is that right? (2,1) yeah, okay. That’s all I know about my parabolas so far.

The next thing I’m going to try to do is look at the Axis of Symmetry, see if that helps me get any more points. What that will tell me is that this point will be reflected across the Axis of Symmetry and show up there as well. So I think my parabola looks something like this, but in order to be precise, I’m going to plug in another x value just to make sure I have some other points.

I’m going to just plug in let x equal to 0, not zero because I already have here x is 0. I’m going to let x equal to 1, I think it’s going to be something around 2 or 3 here let’s see. If in my original equation x is equal 1, then my function is going to be 1² take away 4 times 1 plus 5, so I’ll have 1 take away 4 plus 5, that’s -3 plus 5 which is 2. That tells me another one of my points is (1,2) that makes sense on my graph. Good there it is (1,2) using symmetry I can get this point reflected as well, and now I know better what my parabola looks like I’m like more confident in my choice.

That’s one way to check your work, to pick another x value to substitute in. An alternative way to check your work would be to get a graphing calculator and find some more values. So this problem was a little bit tricky because there were no x intercepts. I found that using the discriminant and also because it couldn’t be factored, you might have found it by using the quadratic equation. However you do it, however you find out there’s no x intercepts, it’s a little bit challenging, just remember to use the Axis of Symmetry and maybe choose another x value to help you complete the shape of your parabola.

## Comments (0)

Please Sign in or Sign up to add your comment.

## ·

Delete