Find the equation of the tangent to the curve $y=(x+3)(x^2-1)$ at the point $(2,15)$.

Question

Find the equation of the tangent to the curve $y = (x+3)(x^2-1)$ at the point $(2,15)$, using the Product Rule.

Answer 1

Given the product rule $f'(x)=(x+3)(2x)+(x^2-1)(1)\Rightarrow f'(x)=3x^2+6x-1$

To find slope $m$ of the tangent line at point $(2,15)$, set $x=2$ in $f'(x)$. Thus, $$m\Rightarrow f'(2)=3(2)^2+6(2)-1\Rightarrow m=23$$ Now we have values for $m, y_1, x_1$. Thus the equation shall be: $$y-y_1=m(x-x_1)$$ $$y-15=23(x-2)$$ $$\therefore y=23x-31$$