Answer
The equation of the tangent line at the given point is $y=4x+10$
Work Step by Step
$f(x)=(x+2)(x^{2}+5)$ $,$ $(-1,6)$
First, use the product rule to evaluate the derivative of the function:
$f'(x)=(x+2)(x^{2}+5)'+(x^{2}+5)(x+2)'=...$
Evaluate the indicated derivatives and simplify:
$...=(x+2)(2x)+(x^{2}+5)(1)=2x^{2}+4x+x^{2}+5=...$
$...=3x^{2}+4x+5$
Substitute $x$ by $-1$ in the derivative found to obtain the slope of the tangent line:
$m=3(-1)^{2}+4(-1)+5=3(1)-4+5=3-4+5=4$
Both the slope of the tangent line and a point through which it passes are now known. Use the point-slope formula of the equation of a line, which is $y-y_{1}=m(x-x_{1})$ to obtain the equation of the tangent line at the given point:
$y-6=4[x-(-1)]$
$y-6=4(x+1)$
$y-6=4x+4$
$y=4x+4+6$
$y=4x+10$