Question

Find the equation of the tangent line to the curve at the given point. y = sqrt(x) (9,3). I understand the derivative of f(x) = sqrt(x), I'm just getting lost when trying to work with the points that aren't (1,1)

Answer 1

So you found the derivative function ok? :)


The equation of the line tangent to our function at (9,3) is given by:
\[\Large y=mx+b\]Where \(\Large m\), the slope, is \(\Large f'(9)\)

Answer 2

Yes, the derivative part is easy for me! \[\sqrt{x}\] = 1/2 x.

But I guess it's the rest that I'm still not understanding.

Answer 3

Hmm there should be a square root in your derivative somewhere :D Maybe you just wrote it sloppy.
\[\Large f(x)=\sqrt x \qquad\to\qquad f'(x)=\frac{1}{2\sqrt x}\]

Answer 4

`The slope of the tangent line`, \(\large m\), is given by the derivative function `evaluated at that point`.

\[\Large m=f'(9)=\frac{1}{2\sqrt 9}\]

Answer 5

You're trying to find a line, that `touches` the curve at x=9 and has the same slope.
A line is of the form \(\large y=mx+b\).

So this problem involves finding 2 unknowns, m and b.
m is the tricky part.

Answer 6

too much? :c
confusing?
head going to esplode?

Answer 7

I thiiiiink I got it... haha.

Answer 8

:3

Answer 9

Apparently not!
I ended up with

y-9 = 1/6x (x-3) and then simplified. But that's not the answer. Blaaah.

Answer 10

Oh you're using point-slope form? ah i see.

Answer 11

So a line in point slope form:\[\Large y-y_o \quad=\quad m(x-x_o)\]Where the given point was:\[\Large (x_o,\;y_o) \quad=\quad (9,\;3)\]

Answer 12

So you found m to be,\[\Large f'(9)\quad=\quad \frac{1}{6}\]

Answer 13

Looks like you were very close.
Just one silly little mistake in the middle there :o

Answer 14

I did it backwards! That's the problem! Yes, 1/6 = m. so the answer is 1/6x + 3/2. Thank you!

Answer 15

yay good job \c:/

Answer 16

Thank you very much for your help, a lot more clear now!