Question

question
find the equation of the line normal to the graph of $f(x)=2x^{3}+2x - 2$ at $x = 1$.
please give your answer in slope - intercept form, $y=mx + b$, or if applicable, in the form $x=a$.
provide your answer below.

Explanation:

Step1: Find the derivative of $f(x)$

First, find the derivative of $f(x)=2x^{3}+2x - 2$. Using the power - rule $(x^n)'=nx^{n - 1}$, we have $f'(x)=6x^{2}+2$.

Step2: Evaluate the derivative at $x = 1$

Substitute $x = 1$ into $f'(x)$. So $f'(1)=6(1)^{2}+2=6 + 2=8$. This is the slope of the tangent line at $x = 1$.

Step3: Find the slope of the normal line

The slope of the normal line $m_n$ and the slope of the tangent line $m_t$ are related by $m_n=-\frac{1}{m_t}$. Since $m_t = 8$, then $m_n=-\frac{1}{8}$.

Step4: Find the point on the curve

Find $y$ - value when $x = 1$ for $y=f(x)$. $f(1)=2(1)^{3}+2(1)-2=2$. So the point on the curve is $(1,2)$.

Step5: Use the point - slope form to find the equation of the normal line

The point - slope form is $y - y_1=m(x - x_1)$. Substitute $(x_1,y_1)=(1,2)$ and $m =-\frac{1}{8}$ into it: $y - 2=-\frac{1}{8}(x - 1)$.

Step6: Rewrite in slope - intercept form

Expand and simplify: $y-2=-\frac{1}{8}x+\frac{1}{8}$, then $y=-\frac{1}{8}x+\frac{1}{8}+2=-\frac{1}{8}x+\frac{1 + 16}{8}=-\frac{1}{8}x+\frac{17}{8}$.

Answer:

$y=-\frac{1}{8}x+\frac{17}{8}$