*y*=

*ax*

^{2}. 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*=

*ax*

^{2}

8 =

*a*(4)^{2}8 = 16

*a*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****.**

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

ReplyDeleteYou'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.

Deletehmmmm okay i guess thank u.

Deletehmmm i guess. okay thank u.

ReplyDelete