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Graphing a Horizontal Parabola - Problem 1 3,098 views

Teacher/Instructor Carl Horowitz
Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level 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 sideways parabola that is in standard form. So just like any parabola that's in standard form, the first thing we need to do is somehow find what the vertex is and there's two ways of doing that. We could either do the -b over 2a or complete the square and I'm going to go ahead and complete the square, so we get another try at interpreting how to read a sideways parabola in standard form.

So complete the square for something like this is exactly the same as if there were xs involved. The first thing we want to do is make our coefficient on our squared term 1, so we want to factor out a negative 1/2 in this case, then we want to make sure we put the coefficient on our y term that when we multiply these two things together we end up with the coefficient we started with, so in this case we want to have a -4 because -4 times -1/2 is 2.

And the last thing we do is bring this plus constant term to the outside. If you wanted to you could subtract it over to the other side, that's another way of doing it either one is perfectly fine.

So now what we want to do is write this term as a perfect square, so we have our -1/2 and we have y something squared and we need to write in something here so that this is a perfect square. Dividing -4 by 2, -2 that's what goes in this spot, -2² what we put here squared goes over here, so this turns into +4. So now we need to make sure we keep our equation balanced.

What I just did was added 4 inside the parenthesis, but when I distribute this negative 1/2 in, what I've actually done is subtracted 2, -1/2 times 4 is -2, so in order to keep it balanced, what I have to do is then add 2 to the outside. I subtracted 2 from this phrase, I need to add 2 to balance it out. If you brought the one over, you would have to subtract 2 from this side as well because whatever you do to one side, you have to do to the other, or you have to do the opposite to the same side.

So what we end up now is x is equal to -1/2, y minus 2² plus 3. So what we have done is we have turned our quadratic equation in standard form in vertex form. Now all we have to do is graph this, so we need to interpret what exactly all this stuff means 1, 2, 3. So thinking to what we know about standard form for a vertical parabola, the +3 typically moves it up the y axis, the difference now is the +3 moves it up the x axis, so instead of going up 3 on the y, we go up 3 on the x, so this is going to go forwards three units.

The y minus 2, if this was x minus 2, it would be going forward two units on the x axis. We've changed our y so now we're going forward two units on the y, forward on the y axis is up, so this is going to go up 2, so I know then that I have gone back three units for my vertex and up two units, so vertex is at the point 3,2.

I know that the negative is going to flip my parabola, normally a x equals y² parabola is going to open up to the positive x axis, the negative is going to flip it, so now it's going to open up this way and the 1/2 is going to make it more gradual, which means it's going to be wider.

You could plug in a point if we wanted to find the exact precise picture of this, I'm just trying to get a sketch, so I know that the graph is going to look something like this, not exact, but it's a rough idea of what this looks like.

So by completing the square, I was able to graph a sideways parabola basically using the same sort of rules we learnt for a vertical line, but just translating it for the x is equals to y² parabola.

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