question find the equation of the line normal to the graph of f(x)=-x³ + 4x² - 1 at x = 1. please give your…

Explanation:

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

Using the power - rule $(x^n)'=nx^{n - 1}$, we have $f(x)=-x^{3}+4x^{2}-1$, so $f'(x)=-3x^{2}+8x$.

Step2: Evaluate the derivative at $x = 1$

Substitute $x = 1$ into $f'(x)$: $f'(1)=-3(1)^{2}+8(1)=-3 + 8=5$.

Step3: Find the slope of the normal line

The slope of the normal line $m_n$ is the negative reciprocal of the slope of the tangent line. Since the slope of the tangent line at $x = 1$ is $m_t = 5$, then $m_n=-\frac{1}{5}$.

Step4: Find the point on the curve at $x = 1$

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

Step5: 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 $m=-\frac{1}{5}$, we get $y - 2=-\frac{1}{5}(x - 1)$. Expand to slope - intercept form: [ \begin{align*} y-2&=-\frac{1}{5}x+\frac{1}{5}\ y&=-\frac{1}{5}x+\frac{1}{5}+2\ y&=-\frac{1}{5}x+\frac{1 + 10}{5}\ y&=-\frac{1}{5}x+\frac{11}{5} \end{align*} ]

Answer:

$y=-\frac{1}{5}x+\frac{11}{5}$