1. ## Tangent line help..

Find the equations of the 2 tangent lines to the graph of f that pass through the indicated point.

f(x)= 4x-x^2 point is (2, 5)

The farthest I have gotten with this is writting this:

(a^2-5)/(a-2)

The next thing is I'm trying to figure out the equal thingy..

a^2-5 = ???

2. Originally Posted by elpermic
Find the equations of the 2 tangent lines to the graph of f that pass through the indicated point.

f(x)= 4x-x^2 point is (2, 5)

The farthest I have gotten with this is writting this:

(a^2-5)/(a-2)

The next thing is I'm trying to figure out the equal thingy..

a^2-5 = ???
What you need to do is first find $f'(x)$:

$f'(x)=4-2x$

Thus, the slope at the point (2,5) is 4-2(2)=0.

So the tangent line is horizontal.

The equation of the tangent would then be $y-5=0(x-2)\implies \color{red}\boxed{y=5}$

Does this make sense?

--Chris

3. Nope.. Actually, the line isn't horizontal at all..

I got a^2-5/a-2 and took the derivative and that is 2a..

Now what do I do?

4. Actually, scratch that, I can't believe how stupid I am.

5. Originally Posted by elpermic
Nope.. Actually, the line isn't horizontal at all.
I think I see what may need to be done...

Let $(x_0,f(x_0))$ be a point on the function.

Then, $f'(x)=4-2x\implies f'(x_0)=4-2x_0$

Thus, the tangent line has the form $y-f(x_0)=f'(x_0)(x-x_0)$

So we see that $y-(4x_0-x_0^2)=(4-2x_0)(x-x_0)$

This line passes through the point $(2, 5)$, so our equation becomes:

$5-(4x_0-x_0^2)=(4-2x_0)(2-x_0)$

So we get $x_0^2-4x_0+5=8-8x_0+2x_0^2$

Getting all the terms on side, we see that $x_0^2-4x_0+3=0$

So $x_0^2-4x_0+3=0\implies (x_0-3)(x_0-1)=0\implies \color{red}\boxed{x_0=1~or~x_0=3}$

Thus, $f(1)=3$ and $f(3)=3$

Therefore, we get two different tangent lines:

$y=2(x-1)+3\implies \color{red}\boxed{y=2x+1}$

$y=-2(x-3)+3\implies \color{red}\boxed{y=-2x+9}$

Does this make sense?

The graph verifies this:

--Chris

6. Yup, those 2 are the answers..

The stupid thing is though, I just had to take the slope from (2,5) and the other 2 points on the graph, and I would've gotten the answer..

The question would've only taken me a minute, and it ends up taking 20 minutes trying to find stuff, lol.