Transforming Parabolas


Stacy Musgrave

We will explore transformations of the following parabola:


Before beginning our transformations, we will put the equation of our parabola into standard form:

From here, we identify the vertex of our parabola at (-3/4, -41/8) and the +2 indicates our parabola opens upward.

To shift the graph to the right by 4, we can replace x with (x-4) everywhere in our original equation (or equivalently in the standard form of our equation).

We can likewise move the vertex into the second quadrant in several ways**. To preserve the shape and direction of our parabola, the transformation we seek is to shift the graph up a distance strictly greater than 41/8. For example, we could add 6 to our equation and get the following:

So the new vertex occurs at (-3/4, -41/8 + 6) = (-3/4, 7/8) in the second quadrant.

Another transformation we could consider is making the original parabola concave down. To do this, we need to change the coefficient of the x^2 term to be negative.

And we see the graphs plotted:


A General Parabola

More generally, we can look at the equation of a parabola
So the vertex occurs at and if a>0, the graph is concave up, and if a<0, the graph is concave down.

How do we verify this information about concavity? We can clearly observe the trend by comparing lots of different graphs, or we could call upon our knowledge of calculus and the second derivative. Since the second derivative of our general parabola formula is y" = 2a, which is positive when a>0 and negative when a<0, we get the above results about concavity.

Now our ability to shift our graph left and right or up and down is directly linked to the vertex. To move the graph around to have the new vertex (h,k), we can solve the following equations to figure out how much to add in the x- and y-directions, respectively (where x and y represent how much to add).

**Returning to Shift to Different Quadrant

In the above discussion of our parabola , I noted that there were several ways to get the parabola's vertex into the second quadrant. The other method I failed to mention earlier involves a reflection across the x-axis. Note this method not only moves the vertex, but also fails to keep the original shape of the graph.

Our new, reflected parabola has it's vertex in the second quadrant, but is now concave down.

We could likewise reflect across the y-axis by replacing every x in the orginal with (-x) to move the vertex into the fourth quadrant.

It should be clear that doing a combination of the previous two reflections can help you shift (though probably not preserving concavity) your vertex to any desired quadrant, assuming you begin with your vertex not on an axis.

And if you really want to get crazy, check out this reflection across the line y=x.




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