Question

Consider the function f(x) = (x+2)^2.
1) Find the equation(s) of the line(s) tangent to y=f(x) that pass through (1,1).
I got: y=12(x-1)+1. Correct?

Answer 1

\[\Large f'(x)=2(x+2)\]The slope, \(\Large m\), of the line tangent to our function at (1,1) is given by \(\Large f'(1)\).

\[\Large f'(1)=2(1+2)=6=m\]

Answer 2

So our tangent line equation will be of the form:\[\Large y=6x+b\]

Answer 3

Oh you were using point-slope form I gather?
So we would set it up like this I guess:\[\Large y-1=6(x-1)\]

Answer 4

you forgot to multiply by the derivative of the inside stuff. Wouldn't it be:

\[f1(x)= 2(x+2)(2) = 4(x+2) = 4(3) = 12\]

Answer 5

I think you're right. The (2) is not suppose to be there, right? the derivative of the inside stuff will just be one?

Answer 6

There's a part 2 to this question, which is to tricky.
Q) Is there a point P = (xo,yo) on the line y=x such that there is only one tangent line to the curve y=f(x) passing through P?

Answer 7

Hellllppppppp me.

Answer 8

Ah sorry I ran away for a moment there :(

Answer 9

Yes, the derivative of the inside stuff will just be one.

Answer 10

I don't blame you lol

Answer 11

Second part? Hmmmm lemme think bout it a sec ;o

Answer 12

kk thank-you! Do you know how I can start part 2?

Answer 13

Is there a point P = (xo,yo) on the line y=x such that there is only one tangent line to the curve y=f(x) passing through P?

Answer 14

kk, sure. The Prof said it's tricky because of something to do with the point

Answer 15

Hmm I'm afraid I don't understand the question. :(
This is how I'm interpreting it, but it's probably way off.One tangent line to the curve f(x) passing through P?
No, because y=x doesn't `touch` the function f(x) at any point.