#### Answer

$a.$ The value of the derivative is $f'(2)=14$.
$b.$ The equation is $y=14x-19.$
$c.$ The graph is on the figure below.

#### Work Step by Step

$a.$ Using the definition of the derivative with $a=2$ we have
$$f'(2)=\lim_{h\to0}\frac{f(2+h)-f(2)}{h}=\lim_{h\to0}\frac{5(2+h)^2-6(2+h)+1-(5\cdot2^2-6\cdot2+1)}{h}=\lim_{h\to0}\frac{5(2^2+h^2+4h)-12-6h+1-9}{h}=\lim_{h\to0}\frac{20+5h^2+20h-12-6h+1-9}{h} = \lim_{h\to0}\frac{5h^2+14h}{h}=\lim_{h\to0}(5h+14)=5\cdot0+14=14.$$
$b.$ The equation of the tangent line through the point $(a,f(a))$ is given by $y-f(a)=f'(a)(x-a)$. Using $a=2$, $f(a) =f(2)= 5\cdot2^2-6\cdot2+1=9$ and $f'(a)=f'(2)=14$ we have
$$y-9=14(x-2)\Rightarrow y-9=14x-28$$
which gives
$$y=14x-19.$$
$c.$ The graph is on the figure below. The function is graphed by the solid line and the tangent is dashed.