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Introduction to Parabolas  Problem 2
Carl Horowitz
Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upperlevel math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
Graphing a parabola in vertex form when a lot of transformations are occurring. So what we're looking at here is a graph in vertex form and it's called vertex form because we know what each of these components does. I know what this does, I know what this does and I know what this does and I can take those things and just transform the graph of x² which I know is just a standard parabola.
So order of operations in theory do matter but when you throw them all together actually the most important thing is finding the vertex and finding which direction the graph is facing and so order in that regard doesn't make that much difference.
So looking at inside my parenthesis, I know that that x plus 2 affects if the graph goes side to side and I know I need to look for what value of x makes that 0 which is 2 so this is going to move my graph backwards 2.
In theory we should be doing the 3 first if we go in order of operations, but like I said the order of operations doesn't quite matter as much so I'm actually going to do the +4 first because that makes my life a little bit easier. That +4 on the outside moves the graph up and down. In this case I know that this is going to move the graph up 4.
So I know that my vertex is back 2 and up 4, back 2 and up 4. Looking then at the 3. The negative tells me that the graph is flipped so it normally faces up, the negative makes it face down and the 3 is going to make the graph skinnier, it's going to squish everything together and how specific you want it to be depends on your teacher. Some teachers will just be happy seeing a steep skinny graph drawn, other are going to want points, so I'll go through at least finding one point and then if you need to find another point you can go ahead and do that on your own.
So the easiest way to find a point is to go to one unit to either side of your vertex, so if we have our x coordinate of our vertex at 2, go to 1 the smaller numbers are easier. So then just plug 1 into your equation and see which you end up getting, so we end up with 3, 1 plus 2² plus 4. 1 plus 2 is 1, 1² is 1 times 3 is 3, so we end up with 3 plus 4 is 1.
So when I plugged in 1, I ended up with the point 1. Using the axis of symmetry, I now know that there is a corresponding point one unit over, so using the axis of symmetry, I can just reflect that over and know that if I'm at one unit to the right of the axis of the symmetry, that same point is going to be one unit to the left.
If you need to find another point, if your teacher wanted to say five points for a parabola, plug in zero and see what comes up. There is a quick little trick though to find that first point and there's not really a proper term for it, but you can look at think of the 3 as bizarre slope. It's not a slope because obviously we're not going to end up with any straight line it's a curve, but what that is going to tell you is if you go one unit to the side of the vertex, that's telling you how far you are changing.
So if I go one unit to the right, I know that I have to go down three units. That only works for finding the first point. The second point is not going to be that same 1 over 3 units down because then you would have a line, but to find that first point that is a pretty easy little short cut to find that guy or this guy. And you do need to remember the negative to know which way you're going because obviously if you went down 3 from here, you'd end up some point over here that wouldn't work.
So using the transformations that we know, up, down side to side and our coefficient we're able to fairly easily graph a parabola in vertex form.
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Carl Horowitz
B.S. in Mathematics University of Michigan
He knows how to make difficult math concepts easy for everyone to understand. He speaks at a steady pace and his stepbystep explanations are easy to follow.
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Introduction to Parabolas
Problem 1 9,672 viewsGraph the parabolas:
f(x) = (x − 1)²f(x) = 2x²f(x) = x²f(x) = (x + 3)²f(x) = ½x² 
Introduction to Parabolas
Problem 2 7,289 viewsGraph:
f(x) = 3(x + 2)² + 4 
Introduction to Parabolas
Problem 3 6,334 viewsa) Give any equation for a parabola with vertex (3,2)b) That also passes through (2,4) 
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Problem 4 6,012 viewsDetermine 'a' and 'k' so that (3,2) and (0,11) lie on y = a(x + 2)² + k

Introduction to Parabolas
Problem 5 555 views 
Introduction to Parabolas
Problem 6 602 views 
Introduction to Parabolas
Problem 7 536 views 
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Problem 8 595 views
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