Need help with slope of the tangent, urgent

[thelazyman]

Hi, the question is h-->0 f(x) = root 3 - x at points (-1, 2)

After calculating i got 1/4 as the slope and when i tried making an equation i got x - 4y + 9 = 0 but the book says the answer is x-4y-7 = 0, did I do anything wrong?

 

[stapel]

Hi, the question is h-->0 f(x) = root 3 - x at points (-1, 2)

Do you mean that you need to find the limit, as h goes to zero, of the (square? cube?) root of 3, less x? Are you finding the limit at the values x = -1 and x = 2? Or at the (single) point (x, y) = (-1, 2)?

After calculating i got 1/4 as the slope...

Please reply showing this work.

When you reply, please include clarification of what the actual question is. Thank you.

Eliz.

 

[skeeter]

you sure did, the slope is -1/4 ...

\(\displaystyle \L \frac{d}{dx}(\sqrt{3-x}) = \frac{-1}{2\sqrt{3-x}}\)