 ###### 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

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# Graphing Quadratic Equations - Problem 2

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

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Here I have a parabola that I’m going to want to graph. I know it’s a parabola because it has x² as the highest power on x. So the first thing I’m going to do is come up with some important points to put on my graph and then connect them. I’m going to find the y intercept, the x intercepts, try to find the vertex, the Axis of Symmetry, all that stuff will help me draw the parabola.

First thing for y intercept, you always put in x is equal to 0, if it’s a parabola, if it’s a line, if it’s a cubic anytime you want to find the y intercept, you always substitute in x equals 0. So if x equal 0, that becomes 0, that becomes 0 and y is equal to 7. So I already know one of my points is (0,7), that was quick.

Now I need to find my x intercepts. To find the x intercepts, you let y equal 0 and then you could either factor, use a quadratic equation, you could complete the square or you could take square root both sides if there’s no b term. It’s up to you, you have lots of different strategies. I’m going to try factoring this guy. I like factoring, but it is some guess and check, a lot of students don’t like factoring and they would probably use the quadratic equation. It’s totally up to you.

If I’m going to factor that guy, I’m going to try two numbers that multiply to 4, 4x², 2 monomials that multiply to 4x². Then I need things that multiply to 7. It’s got to be 7 and 1 except for I need them both to be negative so that they have a negative sum. I’m just guessing this is what it might look like. I’m going to go through and foil and just make sure this is the correct factorization of that trinomial. If I Foil, I’ll have 4x² take away 14x take away 2x plus 7, which is indeed what I was looking for that told me I had factored this correctly.

Okay factoring isn’t my end goal here. What I’m trying to find is the x intercepts. Now that I have the factored form, I’m going to set the factored form equal to 0, and then using the zero products property, I can find my x values. Set each one of those factors equal to 0 and then solve for x. X is going to be ½, that’s going to be one of my x intercepts (½,0). The other x intercept is going to be, add 7 to both sides divide by 2. X is going to be (7/2,0). Those are my two x intercepts, and I have my y intercept.

Next thing I’m going to look for after I’ve done that is the vertex. The way to find the x coordinate of the vertex is to do x is equal to –b/2a. You just have to kind of memorize that. Negative my b value would be -(-16) which is like +16 over 2 times my a value. So I have 16 over 8 which is 2. That’s my x coordinate of the vertex. That’s also the equation for my Axis of Symmetry. Remember the Axis of Symmetry is that vertical line that goes right through your vertex, so there’s my x coordinate.

To find my y value I have to plug in 2 back up there for x. Y is going to be 4 times 2² which is 4, take away 16times 2 plus 7, so that’s 16 take away 32 plus 7 and I get my y value is -9. I think let me just double check that on my calculator in a sec. So I wanted to do 16 take away 32 plus 7, -9 yeah, okay. So that tells me my vertex is going to be 2 for x and -9 for y.

Before I put that all on the graph and start using symmetry to get some more points, I want to think about whether my parabola opens up or down and whether it's wide or skinny and the way I do that is by looking at my a, my leading coefficient, in our case it's 4. What that tells me since 4 is positive, it tells me it’s going to open up like that and my vertex will down here. Also that number 4 tells me I’m going to have a pretty skinny parabola. It’s going to be pretty steep as the y numbers are going to increase pretty steeply as the x numbers increase.

Let’s get this all on the graph and see if we can make some more sense of this. So here I go over to the graph, I’m going to start with the y intercept of (0,7) 1, 2, 3, 4, 5, 6, 7, I found my x intercepts were ½ and 7/2, so I’m just going to approximate ½, and 7/2 is the same thing as 3 ½ 1, 2 oops I drew 2 ½ there we go, ½ and 7/2, ½, 1 ½, 2 ½ yeah good okay.

Then I’m looking for my vertex which is going to be smacked up in the middle of my x intercepts. My vertex was (2,-9) so I’m going to go over 2, down 9 1, 2, 3, 4, 4, 5, 6, 7, 8, 9, now I can see that this is a really skinny parabola. It looks kind of funny because all these points are all spread out. What I’m going to be doing is connecting them, but first I want to find the Axis of Symmetry which is that vertical line that passes through the vertex x equals 2, and I’m going to want to use the Axis of Symmetry to help me find another point up here. Since this point on the y axis was 2 squares away, this point is going to be reflected in 2 squares away.

Now I’m ready to draw my skinny, tall parabola, there it goes you can make sound effects if you want to, okay there they are. So there is my parabola, it was skinny and opened up. I found the x intercept, y intercepts, vertex and Axis of Symmetry and then all of those pieces put together helped me come up with this graph of the quadratic equation.