Question

write an equation of the line tangent to the graph of $f(x)=3x^{3}+8x^{2}$ at the point where $x = - 2$.

Explanation:

Step1: Find the derivative of the function

The derivative of $f(x)=3x^{3}+8x^{2}$ using the power - rule $(x^n)' = nx^{n - 1}$ is $f'(x)=9x^{2}+16x$.

Step2: Evaluate the derivative at $x = - 2$

Substitute $x=-2$ into $f'(x)$: $f'(-2)=9(-2)^{2}+16(-2)=9\times4-32 = 36 - 32=4$. The slope $m$ of the tangent line is 4.

Step3: Find the y - coordinate of the point

Substitute $x = - 2$ into $f(x)$: $f(-2)=3(-2)^{3}+8(-2)^{2}=3\times(-8)+8\times4=-24 + 32 = 8$. So the point on the curve is $(-2,8)$.

Step4: Use the point - slope form of a line

The point - slope form is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(-2,8)$ and $m = 4$. So $y - 8=4(x + 2)$.

Step5: Simplify the equation

Expand the right - hand side: $y-8 = 4x+8$. Then $y=4x + 16$.

Answer:

$y = 4x+16$