# Ap Calculus help.

1. Oct 20, 2005

### cdhotfire

This somehow seems easy, but I cannot get the jist of it.

f(x) = x^3 - x^2 - 4x + 4

The point (a,b) is on the graph of f and the line tangent to the graph at (a,b) passes through the point (0,-8) which is not on the graph of f. Find the values of a and b.

okay so, ive spent about an hour just on this part, this problem had 3 parts.

so i though, maybe, set the original with as and bs, so:

b= a^3 - a ^2 - 4a + 4

then maybe take the derivative

b'= 3a^2 - 2a - 4

so maybe

y-8=b'x

from there on it seems to get pretty nasty,:surprised doesnt seem like the type of problem, that should be this nasty.

maybe someone can shed some light on this.

2. Oct 20, 2005

### hotvette

You are actually very close. You should be able to expand your last equation and get it in terms of only a and b. Same with the first one. You end up with 2 equations in 2 unknowns (I think). Hint: the point (a,b) is on the graph of each function.

Last edited: Oct 20, 2005
3. Oct 20, 2005

### hotvette

Also, if you haven't already, it really helps to draw a picture. You don't have to actually plot the function, just draw a function, pick a point, draw a tangent to it, etc.

4. Oct 20, 2005

### cdhotfire

hmmm, so

b = 3a^3 - 2a^2 - 4a - 8
b = a^3 - a^2 - 4a + 4

0=-2a^3 +a^2 +12

yuck, im sure this wasnt suppose to happen. :(

edit: somehow, i think im making it harder than it seems, looks like a very simple problem. :\

Last edited: Oct 20, 2005
5. Oct 20, 2005

### hotvette

Can't you solve the last equation for a? I got it just by staring at the equation for a few minutes.

Last edited: Oct 20, 2005
6. Oct 20, 2005

### cdhotfire

a= (-6/a^2) - (1/2)??????

i dont know, seems like this question could take up a long, long time. this is supposedly an ap free response question. im thinking this question, is a little too long, to be on a timed test.

7. Oct 20, 2005

### hotvette

Much easier than that. Just look at the equation and try to imagine what values of a will satisfy the equation. The first one I tried was the right one.

Last edited: Oct 20, 2005
8. Oct 20, 2005

### cdhotfire

so i put in 2, get 4x^2 + 3x +6, looks like a mr quadradic to me.:grumpy:

Last edited: Oct 20, 2005
9. Oct 20, 2005

### hotvette

$$0=-2a^3 +a^2 +12$$

This is the one you are trying to solve for a. I like the choice you made.

10. Oct 20, 2005

### cdhotfire

so (2,0) is the only answer?

11. Oct 20, 2005

### hotvette

The question was to find a & b, which you just did. Congratulations!

12. Oct 20, 2005

### cdhotfire

http://img.mauj.com/an/b/BowingSmiley.gif [Broken]
thank you. very much. :)
spent 2 hours on this and its late, time for sleep. thanks again.

Last edited by a moderator: May 2, 2017
13. Oct 20, 2005

### hotvette

Sorry I didn't have a more thoughtful answer (I needed my beauty sleep ). Since you wanted (a,b) to satisfy the 2 equations, you could claim that you are done since you found that. But, you are curious (which is good) that there might be more solutions to $0=-2a^3 +a^2 +12$ since it is a cubic. Good point. You have several options. My first choice is to go ahead a plot it (don't think that's against the rules). Sure enough, there is only 1 real root. So, the other 2 must be? Actually, if there were more than 1 real root, that would mean that a line passing through (0,-8) would be tangent to the function at more than 1 point (i.e. multiple a,b combinations). See attached thumnail illustrating the situation.

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Last edited: Oct 20, 2005