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Introduction to Parabolas  Problem 4
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!
We're now going to look at a applied use of parabola transformation. So what we have here is an equation for a parabola, we know that its vertex is shifted backwards 2. This vertex is shifted up k and there's some sort of vertical stretch going on. This a is going to make it skinnier or wider and we're trying to figure out what those two components are so that these two points 3,2 and 0,11 lie on the curve.
What you have to remember is that in order for a point to lie in a curve, it has to make a statement true, so by plugging each point into this equation, we know that both of those statements have to be true, so let's start with that.
Let's plug in the point 3, 2 so we end up with 2 is equal to a 3 plus 2 is 1² plus k, so this gives us 2 is equal to 1² is just 1, so we end up with a plus k. Doing the same thing for the other point, plug that in and we end up with 11 is equal to a, plug in the 0, we end up with 0 plus 2 which is just 2² is 4 and this is equal to plus k.
So we now have 2 equations with two unknowns. It's basically a system of linear equations. Careful to do substitution or elimination. The first thing I see is that I have k by itself in both equations, so I'm just going to solve for k, end up with k is equal to 2 minus a and over here we end up with k is equal to 11 minus 4a. All I have to do is set these equal to each other 2 minus a is equal to 11 minus 4a and solve. I am going to bring the a's over to this side, so we add a we end up with 3a, subtract 11 we end up with 9, so a is equal to 3.
Just like any other system of equations when we find one thing all we have to do is plug it back in to find the other component, so I have a bunch of relationships between a and k namely these two right here, all I have to do is plug that a in, so a is 3, close them here k is equal to 2 minus 3, k is equal to 1.
So by plugging in the points into the equation, we were able to find our coefficient which tells us the steepness of the curve, in this case it's three times steeper and we're also ale to find the vertical shift we were told that the graph is actually shifted down 1. So creating s system of equations to solve for different components of a parabola transformation.
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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 9,693 viewsGraph the parabolas:
f(x) = (x − 1)²f(x) = 2x²f(x) = x²f(x) = (x + 3)²f(x) = ½x² 
Introduction to Parabolas
Problem 2 7,304 viewsGraph:
f(x) = 3(x + 2)² + 4 
Introduction to Parabolas
Problem 3 6,347 viewsa) Give any equation for a parabola with vertex (3,2)b) That also passes through (2,4) 
Introduction to Parabolas
Problem 4 6,025 viewsDetermine 'a' and 'k' so that (3,2) and (0,11) lie on y = a(x + 2)² + k

Introduction to Parabolas
Problem 5 560 views 
Introduction to Parabolas
Problem 6 609 views 
Introduction to Parabolas
Problem 7 541 views 
Introduction to Parabolas
Problem 8 604 views
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