Answer
$y=2x+3$
The graph of the function and the tangent line is shown below:
Work Step by Step
$y=\dfrac{1}{x^{2}},$ at $(-1,1)$
The slope of the tangent line at the point $P(a,f(a))$ is given by $m=\lim_{h\to0}\dfrac{f(a+h)-f(a)}{h}$
In this case, $a=-1$
Find $f(a+h)$ by substituting $x$ by $-1+h$ in $f(x)$ and simplifying:
$f(-1+h)=\dfrac{1}{(-1+h)^{2}}=\dfrac{1}{h^{2}-2h+1}$
Since $f(-1)$ is the $y$-coordinate of the point given, $f(-1)=1$
Substitute the known values into the formula that gives the slope of the tangent line and evaluate:
$m=\lim_{h\to0}\dfrac{\dfrac{1}{h^{2}-2h+1}-1}{h}=\lim_{h\to0}\dfrac{\dfrac{1-(h^{2}-2h+1)}{h^{2}-2h+1}}{h}=...$
$...=\lim_{h\to0}\dfrac{1-h^{2}+2h-1}{h(h^{2}-2h+1)}=\lim_{h\to0}\dfrac{-h^{2}+2h}{h(h^{2}-2h+1)}=...$
$...=\lim_{h\to0}\dfrac{h(-h+2)}{h(h^{2}-2h+1)}=\lim_{h\to0}\dfrac{-h+2}{h^{2}-2h+1}=...$
$...=\dfrac{-(0)+2}{0^{2}-2(0)+1}=\dfrac{2}{1}=2$
The point-slope form of the equation of a line $y-y_{0}=m(x-x_{0})$. Both the slope of the line and a point through which it passes are known. Substitute them into the formula to obtain the equation of the tangent line:
$y-1=2(x+1)$
$y-1=2x+2$
$y=2x+2+1$
$y=2x+3$
