First let's graph it.
Notice how the vertex is located in the third quadrant and it appears to be close to . In order to find the actual location of the vertex, we may take the derivative of the binomial and set it equal to 0.
So we find that . Setting the derivative equal to 0, we get:
We then use the value that we found for x to find the y value of the vertex.
So the vertex is located at.
For the next part, we need to replace each x in the binomial by. So the binomial we get is . Then we graph this new binomial on the same axes as the original binomial and we get:
Notice how the new graph appears to be translated 4 units to the right. By solving for the derivative again, setting it equal to zero, and then using the x-value that we get to find the y-value, we find that the vertex of the new graph is . Notice that the y-value stayed the same, but the x-value increased by 4 units, which shows that the graph was translated 4 units to the right.
Using the same method as before to find the vertex, I find that the vertex is .
Finally we need to change the equation to produce a graph, which is concave down and shares the same vertex as the previous equation. After several attempts, I find that the new equation is and it is graphed below.
To confirm that the new graph shares the same vertex as the previous graph, I use the same method as before. I find that the vertex of the new graph is . We see that the vertices are indeed the same and so I have found a graph that is concave down and shares the same vertex as the previous graph.