Amberly Roberts

**Assignment 2: An Investigation of the parabola represented by the equation**

First, we will generate a graph of this equation.

Next, we will overlay a new graph replacing each x by (x - 4). Here's what the graph looks like.

What happened?

Is it clear that every point on the parabola shifted 4 spaces to the right?

Next, we will change the equation to move the vertex of the original graph (the purple one) into the second quadrant.

From this, we notice that the vertex of our graph is at the point (-3/4, -41/8). If you look back at the graph, this information corresponds to the picture. Whew!

If we want the vertex to be in the second quadrant, we need to pick and ordered pair in which the x-coordinate is negative and the y-coordinate is positive. So, let's be rather simple and choose (-2. 3). Now, go back and replace (h, k) with these new values.

Let's graph this equation to geometrically affirm its correctness.

It worked. The vertex is in the second quadrant.

Now, we want to change the equation to produce a graph concave down that shares the same vertex.

How will we do this?

Well, in our investigation above, it is clear that manipulating the "h" and "k" values simply moves the vertex of the parabola. The only thing we have not manipulated is the "a" value.

Let's make the "a" value negative and see what happens. Here's the graph.

From this investigation, it becomes clear that working with parabolas in vertex form allows up to easily manipulate and make changes algebraically that will provide the graphical changes we want to see.

We can decide where we want the parabola to be placed, and plugging values into the vertex form of the equation allows us to accurately place the parabola.

Is working with the coefficients in general form just as easy?