Answer
$$L = 8\sqrt {65} $$
Work Step by Step
$$\eqalign{
& y = - 8x - 3{\text{ on }}\left[ { - 2,6} \right] \cr
& {\text{Definition of Arc Length for }}y = f\left( x \right): \cr
& {\text{Let }}f{\text{ have a continuous first derivative on the interval }}\left[ {a,b} \right]{\text{ The length of the curve }} \cr
& {\text{from }}\left( {a,f\left( a \right)} \right){\text{ to }}\left( {b,f\left( b \right)} \right){\text{ is }}L = \int_a^b {\sqrt {1 + f'{{\left( x \right)}^2}} } dx \cr
& {\text{Notice that }}y = f\left( x \right) = - 8x - 3{\text{ and }}\left[ { - 2,6} \right] \to a = - 2{\text{ and }}b = 6.{\text{ then}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ { - 8x - 3} \right] \cr
& f'\left( x \right) = - 8 \cr
& {\text{Using the arc length formula}}{\text{, we have}} \cr
& L = \int_{ - 2}^6 {\sqrt {1 + {{\left( { - 8} \right)}^2}} } dx \cr
& {\text{simplifying}} \cr
& L = \int_{ - 2}^6 {\sqrt {65} } dx \cr
& {\text{integrate}} \cr
& L = \sqrt {65} \left( x \right)_{ - 2}^6 \cr
& L = \sqrt {65} \left( {6 - \left( { - 2} \right)} \right) \cr
& L = \sqrt {65} \left( 8 \right) \cr
& L = 8\sqrt {65} \cr} $$
