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Introduction to Parabolas  Problem 3
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!
Finding the equation for a parabola when we are given the vertex. So for this example what we're looking at is give any equation for a parabola with the vertex 3, 2.
The easiest way to do this is to go to vertex form which is f of x is equal to ax minus hÂ² plus k. So this is our vertex form, easy way to figure out the vertex. Remember a is just the steepness it's going to affect the steepness and the direction, h affects the side to side and k affects the up and down.
So by simply plugging in anything for a, this will work because we don't actually this says any equation, a just affects the steepness, there's many equations that can have this point as his vertex it' s actually infinite because we can always change that a to get a different graph where the vertex stays the same.
So the h moves the graph side to side and remember we want to be able to plug the x coordinate in to get this middle term to be 0. So I want to be able to plug 3 in here and I end up with a zero here which means I have to add 3. It's always opposite when it's inside that squared term. And then this is how I want to move the graph up or down, plus moves it up, minus moves it down, I want to end up at 2, so therefore I need to subtract 2.
So this is a equation for a parabola with this vertex, in this one we just have the steepness. I could throw a 2, I could throw a 5, I could throw 172 and what's being asked for would still hold. The vertex would still be in that spot, this would just change the shape which is perfectly fine because it says an equation, so I gave the most simple answer.
The second part of this question is asking for a specific graph. We have the vertex and we also needed to satisfy another point. So to satisfy that I do need to throw that a in there because I need to find the a that affects the steepness that pass through that other points.
So first thing I want to do is throw the a in there and the next thing all we have to do is plug in that point and solve for a. So we're told that x is 2 and y is 4 as on the curve, so let's plug those in. 4 is equal to a, plug in 2 plus 3 is just 1, add the 6 2 around and we end up with the 6 is equal to a.
So just by plugging in that point into the equation, we'll be able to solve for a to figure out the steepness, piecing that all together, what we end up with is the equation for a parabola with the given vertex that also passes through the point, so that was asked for.
Whenever we're given a vertex, it's really easy to find the equation for the parabola in vertex form because all we have to do is plug in some information, you don't have to solve for much unless you're given any other point in which case it you'll make one of your equations pretty easy to solve.
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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 9,819 viewsGraph the parabolas:
f(x) = (x − 1)²f(x) = 2x²f(x) = x²f(x) = (x + 3)²f(x) = ½x² 
Introduction to Parabolas
Problem 2 7,424 viewsGraph:
f(x) = 3(x + 2)² + 4 
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Problem 3 6,449 viewsa) Give any equation for a parabola with vertex (3,2)b) That also passes through (2,4) 
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Problem 4 6,132 viewsDetermine 'a' and 'k' so that (3,2) and (0,11) lie on y = a(x + 2)² + k

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Problem 5 648 views 
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Problem 6 694 views 
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Problem 7 628 views 
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Problem 8 690 views
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