Transformations of Parabolas

by Kassie Smith



First, we will graph the parabola given.

We can find the vertex through a multitude of ways. First, if the reader has Graphing Calculator, he can click on the curve and drag the marker along the curve to find the vertex. Students may also use graphing calculators such as the TI-83, TI-84, and TI-Nspire. Algebra can also be used to find the vertex. Using algebra, students may rewrite the equation in vertex form. Students may use the quadratic formula to solve for the roots. Knowing that the x-coordinate of the vertex must lie exactly between the two roots, the student can find the x-coordinate of the vertex and then plug the x value into the function to get the y-coordinate of the vertex. Even easier, students can just use the axis of symmetry or the first part of the quadratic formula , -b/2a, to get the x-coordinate. Regardless of the method, the vertex is (-.75, -5.125).

Then we can replace each x with (x-4) in the equation given. This results in the following graph.

As you can see, this resulted in a horizontal shift to the right of 4 units. So, the y-coordinate of the vertex remains the same and the x-coordinate is 4 more.


The next part is to complete a transformation so that the vertex is in the second quadrant. There are infinitely many ways to do this transformation. For the sake of this assignment, I will just describe one transformation to move the vertex.

Since the original vertex is in the third quadrant, the x-coordinate is already in a satisfactory position. So, I will just focus on moving the y-coordinate. The y-coordinate must be positive for the vertex to be in the second quadrant. The original y-coordinate of the vertex was -5.125 so I must translate the graph upward by more than than amount. For ease, I just changed the sign of the constant at the end of the equation, effectively adding 8 to the equation. So my new vertex is (-.75, 2.875).


Next, we are charged with the task of finding an equation of a new graph that is concave down and shares the same vertex as the original equation. The first thing I want to do is to make the graph concave down. I will try to do this by negating all the terms with a x in them.

I can see that I am close now. I only need to translate my parabola downward so that the vertex of this newer equation coincides with the vertex of the original equation. I can again use any method I want to find the vertex of this graph (technology, vertex form, substituting the x-value in to equation, etc). The y-coordinate of this vertex is -2.875. So the difference between the two vertices is 5.125-2.875, which is 2.25. So, I need to move my latest parabola down by 2.25 units. Here is the final product.


And we can look at these two equations together.

Lastly, in attempt to make it easier to track the progress, here is one picture that shows each transition.