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

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

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

amistre64 · Answer

theres nothin there

anonymous · Answer

It is y= sqrt x

amistre64 · Answer

the equation of the tangent line then is:

y = f'(9)x -9f'(9) + 3

amistre64 · Answer

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

amistre64 · Answer

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

anonymous · Answer

I have no idea what you did.

anonymous · Answer

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

anonymous · Answer

Ah.. ok

anonymous · Answer

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

anonymous · Answer

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

amistre64 · Answer

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