Answer
$y=-8x+9$
The graph of the function and the tangent line is shown below:
Work Step by Step
$f(x)=-2x^{2}+1,$ at $(2,-7)$
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, $a=2$
Find $f(a+h)$ by substituting $x$ by $2+h$ in $f(x)$ and simplifying:
$f(2+h)=-2(2+h)^{2}+1=-2(4+4h+h^{2})+1=...$
$...=-8-8h-2h^{2}+1=-2h^{2}-8h-7$
Since $f(a)$ is the $y$-coordinate of the point given, $f(a)=-7$
Substitute the known values into the formula that gives the slope of the tangent line and evaluate:
$m=\lim_{h\to0}\dfrac{(-2h^{2}-8h-7)+7}{h}=\lim_{h\to0}\dfrac{-2h^{2}-8h}{h}=...$
$...=\lim_{h\to0}\dfrac{h(-2h-8)}{h}=\lim_{h\to0}(-2h-8)=...$
$...=-2(0)-8=-8$
The point-slope form of the equation of a line is $y-y_{0}=m(x-x_{0})$. Both the slope of the line and a point through which it passes are given. Substitute them into the formula to obtain the equation of the tangent line:
$y+7=-8(x-2)$
$y+7=-8x+16$
$y=-8x+16-7$
$y=-8x+9$
