Answer
$a.$ the slope is $m_{tan} = -11.$
$b.$ the equation is $y=-11x+4.$
$c.$ the graph is on the figure below.
Work Step by Step
$a.$ Using the expression for the slope of the tangent line given in definition (1) with $a=1$ (the first coordinate of the point $P(1,-7)$) we have
$$m_{tan} = \lim_{x\to 1}\frac{f(x)-f(1)}{x-1} = \lim_{x\to 1}\frac{-3x^2-5x+1 - (-3\cdot1^2-5\cdot1+1)}{x-1} =\lim_{x\to 1}\frac{-3x^2-5x+1 +7}{x-1} = \lim_{x\to 1}\frac{-3x^2-5x+8}{x-1}.$$
To find this limit we will factorize the quadratic function in the numerator. For its zeros by the formula for zeros of the quadratic function we have
$$x_{1,2}=\frac{-(-5)\pm\sqrt{(-5)^2-4\cdot(-3)\cdot8}}{2\times(-3)} = \frac{5\pm 11}{-6} = \left\{^{-8/3}_1\right.$$
Now we know that every quadratic function of the form of $a_1x^2+a_2x+a_3$ with zeros $x_1$ and $x_2$ can be factored as
$$a_1x^2+a_2x+a_3=a_1(x-x_1)(x-x_2)$$ which in our case gives
$$-3x^2-5x+8 = -3(x+8/3)(x-1) = -(3x+8)(x-1).$$
Now returning to solving the limit we have
$$m_{tan} = \lim_{x\to 1}\frac{-3x^2-5x+8}{x-1} =\lim_{x\to 1} \frac{-(3x+8)(x-1)}{x-1} = \lim_{x\to 1}(-(3x+8))=-11.$$
$b.$ Using the formula fro the equation of the tangent $y-f(a)=m_{tan}(x-a)$ for $a=1$ and $f(a)=-7$ (the second coordinate of point $P(1,-7)$) we have
$$y-(-7) = -11(x-1)\Rightarrow y+7=-11x+11$$
which gives $$y=-11x+4.$$
$c.$ The graph is on the figure below. The function is solid and the tangent is dashed.
