Question

Find an equation of the tangent line to the curve at the point (9, 3).

http://www.webassign.net/wastatic/watex/img/sqrt1a.gif

Answer 1

theres nothin there

Answer 2

It is y= sqrt x

Answer 3

the equation of the tangent line then is:

y = f'(9)x -9f'(9) + 3

Answer 4

f'(x) = \(\frac{1}{2\sqrt{x}}\)

Answer 5

y = 1/6 x -9(1/6) + 3(2/2)
-3/2 +6/2
y = (1/6) x + (3/2)

Answer 6

I have no idea what you did.

Answer 7

1. take the derivative
2. replace x by 9 that is your slope
3. use point slope formula to find the equation of the line

Answer 8

Ah.. ok

Answer 9

\[f(x)=\sqrt{x}\]
\[f'(x)=\frac{1}{2\sqrt{x}}\]
\[f'(9)=\frac{1}{2\sqrt{9}}=\frac{1}{6}\]

Answer 10

\[y-y_1=m(x-x_1)\]
\[y-3=\frac{1}{6}(x-9)\]
\[y=\frac{1}{6}(x-9)+3\]

Answer 11

hey sat; you got a groupie calling for you ;)