find an equation of the normal line to the parabola y = x² - 6x + 3 that is parallel to the line x - 4y = 9…

Explanation:

Step1: Find the slope of the given line

Rewrite $x - 4y=9$ in slope - intercept form $y = mx + b$. We get $y=\frac{1}{4}x-\frac{9}{4}$, so the slope of the line is $m_1=\frac{1}{4}$.

Step2: Find the derivative of the parabola function

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

Step3: Find the slope of the tangent line

Since the normal line is perpendicular to the tangent line, and the slope of the normal line is $\frac{1}{4}$ (parallel to the given line), the slope of the tangent line $m_2$ satisfies $m_1m_2=-1$. So $m_2=-4$.

Step4: Solve for x - values on the parabola

Set $y' = 2x-6=-4$. Solving for $x$: [ \begin{align*} 2x-6&=-4\ 2x&=2\ x&=1 \end{align*} ]

Step5: Find the y - value on the parabola

When $x = 1$, $y=1^{2}-6\times1 + 3=1 - 6 + 3=-2$.

Step6: Find the equation of the normal line

Using the point - slope form $y - y_1=m(x - x_1)$ with $(x_1,y_1)=(1,-2)$ and $m=\frac{1}{4}$, we have $y+2=\frac{1}{4}(x - 1)$. Expanding gives $y=\frac{1}{4}x-\frac{1}{4}-2=\frac{1}{4}x-\frac{9}{4}$.

Answer:

$y=\frac{1}{4}x-\frac{9}{4}$