Question

for each problem, find the equation of the line normal to the function at the given point. if the normal line is a vertical line, indicate so. otherwise, your answer should be in slope - intecept form. 5) $y = x^{3}-3x^{2}+2$ at $(0,2)$

Explanation:

Step1: Find the derivative of the function

Differentiate $y = x^{3}-3x^{2}+2$ using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$.
$y'=\frac{d}{dx}(x^{3}-3x^{2}+2)=3x^{2}-6x$

Step2: Evaluate the derivative at the given point

Substitute $x = 0$ into $y'$.
$y'(0)=3(0)^{2}-6(0)=0$

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 $m_t = y'(0)=0$, the slope of the normal line is undefined (because $m_n=-\frac{1}{0}$), which means the normal line is a vertical line.

Answer:

The normal line is a vertical line.