Answer
$y=-8x-7$
The graph of the function and the tangent line is shown below:
Work Step by Step
$f(x)=4x^{2}-3,$ 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:
$f(-1+h)=4(-1+h)^{2}-3=4(1-2h+h^{2})-3=...$
$...=4-8h+4h^{2}-3=4h^{2}-8h+1$
Since $f(a)$ is the $y$-coordinate of the point given, $f(a)=1$
Substitute the known values into the formula that gives the slope of the tangent line and evaluate:
$m=\lim_{h\to0}\dfrac{(4h^{2}-8h+1)-1}{h}=\lim_{h\to0}\dfrac{4h^{2}-8h}{h}=...$
$...=\lim_{h\to0}\dfrac{h(4h-8)}{h}=\lim_{h\to0}(4h-8)=4(0)-8=-8$
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:
$y-1=-8(x+1)$
$y-1=-8x-8$
$y=-8x-8+1$
$y=-8x-7$
