#### Answer

$a.$ The value of the derivative is $f'(1)=8$.
$b.$ The equation is $y=8x-13$.
$c.$ The graph is on the figure below.

#### Work Step by Step

$a.$ Using the definition of the derivative with $a=1$ we have
$$f'(1)=\lim_{h\to0}\frac{f(1+h)-f(1)}{h}=\lim_{h\to0}\frac{3(1+h)^2+2(1+h)-10-(3\cdot 1^2+2\cdot1-10)}{h}=\lim_{h\to0}\frac{3(1+h^2+2h)+2+2h-10-3-2+10}{h}=\lim_{h\to0}\frac{3+3h^2+6h+2+2h-10-3-2+10}{h} = \lim_{h\to0}\frac{3h^2+8h}{h}=\lim_{h\to0}(3h+8)=3\cdot0+8=8.$$
$b.$ The equation of the tangent line through the point $(a,f(a))$ is given by $y-f(a)=f'(a)(x-a)$. Using $a=1$, $f(a) =f(1)= 3\cdot1^1+2\cdot1-10=-5$ and $f'(a)=f'(1)=8$ we have
$$y-(-5)=8(x-1)\Rightarrow y+5=8x-8$$
which gives
$$y=8x-13.$$
$c.$ The graph is on the figure below. The function is graphed by the solid line and the tangent is dashed.