Test 7 Section 3 - #18 (page 773)

This one's a huge pain, and before I get going on it I should stress that it's extremely unlikely that you will see something like this on your SAT...so don't spend too much time worrying about it.  If you want to worry anyway...read on!


We know two points: (0, 6) (the ball was 6 feet high at time 0 when it was thrown), and (2.5, 106) (that's the time and max height they give us). We can plug both of those points into the function to figure out the constants. Start with (0, 6):

6 = c - (d - 0)²

6 = c - d²

Ok, now put in (2.5, 106):

106 = c - (d - 4(2.5))²

106 = c - (d - 10)²

FOIL it out:

106 = c - (d² - 20d + 100) 

Distribute the negative:

106 = c - d² + 20d - 100

Ok here's where it gets crazy. We know that c - d² = 6 (I bet you were wondering why I made it red...), so substitute that into the 2nd equation

106 = 6 + 20d - 100

100 = 20d - 100

200 = 20d

10 = d

now use c - d² = 6 one more time to figure out c:

c - 10² = 6

c - 100 = 6

c = 106

OK! NOW, we can figure out what's going on at time  t = 1:

h(t) = c - (d - 4t)²

h(1) = 106 - (10 - 4)²

h(1) = 106 - 6²

h(1) = 106 - 36

h(1) = 70

So there, our answer, finally, is 70.

Note that there's a shortcut here, but it's very tricky to see it.  The key is to recognize that this equation will hit its maximum (remember, the maximum was at t = 2.5) when (d - 4t)² = 0.  From there, you know immediately that c = 106, and d = 10.  Want more clarification on this shortcut?  Ask in the comments!