Answer
$m=-4$
Work Step by Step
$f(x)=1+2x-3x^{2},$ at $(1,0)$
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:
$f(1+h)=1+2(1+h)-3(1+h)^{2}=...$
$...=1+2+2h-3(1+2h+h^{2})=...$
$...=1+2+2h-3-6h-3h^{2}=-3h^{2}-4h$
Since $f(a)$ is the $y$-coordinate of the point given, $f(a)=0$
Substitute the known values into the formula that gives the slope of the tangent line and evaluate:
$m=\lim_{h\to0}\dfrac{-3h^{2}-4h-0}{h}=\lim_{h\to0}\dfrac{-3h^{2}-4h}{h}=...$
$...=\lim_{h\to0}\dfrac{h(-3h-4)}{h}=\lim_{h\to0}(-3h-4)=...$
$...=-3(0)-4=-4$
