Math Help - linear approximation y=

1. linear approximation y=

problem follows, "let y be the linearisation of the function $f(x) = \frac{1}{x}$ at $x = 2$.

then y=?

they give the answer to be $y = - \frac{1}{4}x + 1$

I get this to be the answer
$
\begin{array}{l}
y = f'(a)(x - a) + f(a) \\
= - \frac{1}{4}x - 2 + \frac{1}{2} \\
= - \frac{1}{4}x - 1.5 \\
\end{array}
$

Have I done something wrong, and if so where?

2. Yes, you've made an error; it's in your algebra:
$y=f'(a)(x-a)+f(a)$
$y=-\frac{1}{4}(x-2)+\frac{1}{2}$
$y=-\frac{1}{4}\cdot{x}-\left(-\frac{1}{4}\right)\cdot(2)+\frac{1}{2}$
$y=-\frac{1}{4}\cdot{x}+\frac{1}{2}+\frac{1}{2}$
$y=-\frac{1}{4}\cdot{x}+1$

--Kevin C.

3. ah bugger. Thinking I should have had more sleep.
Thanks