Question

find an equation for the tangent line to the curve at the given point. then sketch the curve and tangent line together.
y = 2x^3, (1,2)
an equation for the tangent line is y - 2 = 3x - 3 (type an equation.)

Explanation:

Step1: Find the derivative of the function

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

Step2: Evaluate the derivative at the given point

Substitute $x = 1$ into $y'$. When $x = 1$, $y'(1)=6\times1^{2}=6$. This is the slope $m$ of the tangent line.

Step3: Use the point - slope form of a line

The point - slope form is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(1,2)$ and $m = 6$. So $y - 2=6(x - 1)=6x-6$.

Answer:

$y - 2=6x - 6$