Answer
Tangent Line: $y=3x-8$
Work Step by Step
$f(x)=(x-2)(x^{2}+4), (1,-5)$
a. Use the Product Rule to Find f'(x): $f'(x)=f_{1}(x)f_{2}'(x)+f_{2}(x)f_{1}'(x)$
$f'(x)=(x-2)(2x)+(x^{2}+4)(1)$
$f'(x)=(2x^{2}-4x)+(x^{2}+4)$
Plug in Given x Value to Find Slope: $f'(1)=(2[1]^{2}-4[1])+([1]^{2}+4)=(2-4)+(1+4)=3$
Plug Found Slope and Given Point Values into Point Slope Form: $y-y_{1}=(slope)(x-x_{1})$
$y-[-5]=3(x-[1])$
$y+5=3x-3$
Equation of Tangent Line: $y=3x-8$
b. Can be found using a graphing calculator
c. This answer was confirmed using a graphing calculator
