Answer
$m=-11$
Work Step by Step
$f(x)=4x^{2}-3x,$ at $(-1,7)$
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, $a=-1$
Find $f(a+h)$ by substituting $x$ by $-1+h$ in $f(x)$ and simplifying:
$f(-1+h)=4(-1+h)^{2}-3(-1+h)=...$
$...=4(1-2h+h^{2})+3-3h=4-8h+4h^{2}+3-3h=...$
$...=4h^{2}-11h+7$
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:
$m=\lim_{h\to0}\dfrac{(4h^{2}-11h+7)-7}{h}=\lim_{h\to0}\dfrac{4h^{2}-11h}{h}=...$
$...=\lim_{h\to0}\dfrac{h(4h-11)}{h}=\lim_{h\to0}(4h-11)=...$
$...=4(0)-11=-11$
