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Introduction to Parabolas  Problem 1
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!
Transformations of a parabola are similar to transformations of any other function that you've looked at. So whenever we are dealing with x² plus or minus a number, the number isn't directly associated with the exponent, we know that this is going to be a vertical shift.
So in this case my vertical shift is up 2 which takes my vertex from the origin and moves it up two units, so my vertex is now at y equals 2. My axis of symmetry is always the line straight up and down from our vertex, so our axis of symmetry hasn't changed with this transformation.
X² minus 1 is very, very similar except instead of moving it up, we're moving it down, so we move it down one unit, the graph of the parabola pretty much stays the same, our axis of symmetry doesn't move either. So that is if our number is on the outside of the function.
We also can have it where the number is associated directly with the x, what I would call inside the function, and what this does is it affects the x value. The best way I think to think of this is to think about what value of x will make this 0? In this case x is equal to 1 and that is going to be our x value for our vertex. So I know I go over one unit and there I have my parabola.
In this case the axis of symmetry has moved because it goes along with the x coordinate of the vertex, so now my axis of symmetry has moved over to 1 as well.
Very similarly x plus 3², the 3 is associated with the x, so that tells me it's a side to side transformation what value of x makes this 0? 3 so I go back three units 1, 2, 3 and once again our axis of symmetry has moved as well.
So what happens when we have a coefficient? Coefficient is going to multiply the y value, so when we have a coefficient of a number bigger than 1, what happens is our graph becomes steeper, when we have a coefficient less than 1, our graph becomes wider and the way you can think of that is we're basically just multiplying the y value by the number in front. So when we have the y value of 0,the graph doesn't change. Before if x was 1, we have the point 1,1 and now when x is 1, we have the point y is 2, so we took our old y value and multiplied it by 2, that's going to cover for every single y value so it's graph is just basically going to get squished together and become significantly steeper. There is no transformation side to side, so the axis of symmetry hasn't moved.
By similar logic this 1/2 is going to multiply the ys by 1/2, so when y is 0, our point doesn't change, but before we add the point 1, 1 we now have the point 1, 1/2 and so what's happened here is the graph just becomes a lot wider, axis of symmetry doesn't change because your vertex hasn't moved.
The last thing that's going to happen is a negative coefficient and basically what that does is it flips the graph down. So the coefficient if it's a positive coefficient it's just going to look similar to how it normally looks, if it's a negative coefficient it flips the graph down and so we end up with the same shape as normal, but just looking down. If it was say like a 2, it would be skinnier and facing down, a combination of the two.
So in short what we have is something that looks like this. F(x) is equal to a quantity x minus 8 squared plus k. This is called vertex form because from this form we can find the vertex. The k is going to affect the shifting up and down, the h is going to affect the shifting side to side and the a is that coefficient which dictates how steep or wide it is or if it's facing up or facing down.
So transformations broken down one by one, we can do them more than one at a time by using this formula, but it's really important to understand what each variable in this formula does.
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Carl Horowitz
B.S. in Mathematics University of Michigan
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Introduction to Parabolas
Problem 1 9,371 viewsGraph the parabolas:
f(x) = (x − 1)²f(x) = 2x²f(x) = x²f(x) = (x + 3)²f(x) = ½x² 
Introduction to Parabolas
Problem 2 7,099 viewsGraph:
f(x) = 3(x + 2)² + 4 
Introduction to Parabolas
Problem 3 6,149 viewsa) Give any equation for a parabola with vertex (3,2)b) That also passes through (2,4) 
Introduction to Parabolas
Problem 4 5,834 viewsDetermine 'a' and 'k' so that (3,2) and (0,11) lie on y = a(x + 2)² + k

Introduction to Parabolas
Problem 5 425 views 
Introduction to Parabolas
Problem 6 472 views 
Introduction to Parabolas
Problem 7 420 views 
Introduction to Parabolas
Problem 8 442 views
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