Question

question
find the equation of the line tangent to the graph of $f(x)=2x^{3}-2x^{2}$ at $x = - 1$.
provide your answer below.
$y=square$

Explanation:

Step1: Find the derivative of the function

The derivative of $f(x)=2x^{3}-2x^{2}$ using the power - rule $(x^n)' = nx^{n - 1}$ is $f'(x)=6x^{2}-4x$.

Step2: Evaluate the derivative at $x = - 1$

Substitute $x=-1$ into $f'(x)$: $f'(-1)=6(-1)^{2}-4(-1)=6 + 4=10$. This is the slope $m$ of the tangent line.

Step3: Find the y - coordinate of the point of tangency

Substitute $x = - 1$ into the original function $f(x)$: $f(-1)=2(-1)^{3}-2(-1)^{2}=2(-1)-2(1)=-2 - 2=-4$. So the point of tangency is $(-1,-4)$.

Step4: Use the point - slope form of a line

The point - slope form is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(-1,-4)$ and $m = 10$. So $y+4 = 10(x + 1)$.

Step5: Simplify the equation

Expand and simplify: $y+4=10x + 10$, then $y=10x+6$.

Answer:

$y = 10x+6$