Ap Calculus help.

cdhotfire · Oct 20, 2005

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.

thanks in advanced.:tongue2:

hotvette · Oct 20, 2005

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.

hotvette · Oct 20, 2005

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.

cdhotfire · Oct 20, 2005

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. :\

hotvette · Oct 20, 2005

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

cdhotfire · Oct 20, 2005

hotvette said:

Can't you solve the last equation for a? Cubics are rarely easy.

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.

hotvette · Oct 20, 2005

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.

cdhotfire · Oct 20, 2005

hotvette said:

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.

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

edit: roar!, mr quad come out a bad choice.
so only one answer? (2,0)?

hotvette · Oct 20, 2005

[tex]0=-2a^3 +a^2 +12[/tex]

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

cdhotfire · Oct 20, 2005

so (2,0) is the only answer?

hotvette · Oct 20, 2005

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

cdhotfire · Oct 20, 2005

hotvette said:

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

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.

hotvette · Oct 20, 2005

cdhotfire said:

so (2,0) is the only answer?

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 [itex]0=-2a^3 +a^2 +12[/itex] 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.

Log in or Sign up

Ap Calculus help.

cdhotfire

hotvette

hotvette

cdhotfire

hotvette

cdhotfire

hotvette

cdhotfire

hotvette

cdhotfire

hotvette

cdhotfire

hotvette

Attached Files:

plot.gif