## 4/28/10

### Test 3 Section 5 - #18 (page 530)

Ok.  They tell us the equation of that parabola is y = ax2.  Whenever you get a question like this (which is not all that uncommon in the hard math questions) you're looking to find a point on the graph!  Since they also tell us that the area of the square is 64, we know the sides of the square must each be 8:

Since parabolas are symmetrical and this one has its minimum at (0,0), we know now what the actual points are at each corner of our square:
And that, friends, is where this gets fun.  Let's use (4,8) since both numbers are positive.  Now that we have a point on our parabola, we just need to plug values in for x and y!

y = ax2
8 = a(4)2
8 = 16a
0.5 = a

This would, of course, also work with (-4,8). Either way, you're going to get 0.5 (or 1/2) for a.

1. Why won't (0,0) work, its also a point on the graph. I commented:)

1. You're trying to solve for a, and you can't do that if you use (0, 0)--all that tells you is that 0 = a(0), which doesn't get you anywhere.

2. hmmmm okay i guess thank u.

2. hmmm i guess. okay thank u.