find the equation of the tangent line to the graph of the equation at (1, 1). f(x)=-6x^4 + 4x^3 - 3x^2 + 5x…

Explanation:

Step1: Find the derivative of the function

Using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, if $f(x)=-6x^{4}+4x^{3}-3x^{2}+5x + 1$, then $f'(x)=-24x^{3}+12x^{2}-6x + 5$.

Step2: Evaluate the derivative at $x = 1$

Substitute $x = 1$ into $f'(x)$: $f'(1)=-24(1)^{3}+12(1)^{2}-6(1)+5=-24 + 12-6 + 5=-13$. The slope $m$ of the tangent line at $(1,1)$ is $-13$.

Step3: Use the point - slope form of a line

The point - slope form is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(1,1)$ and $m=-13$. So $y - 1=-13(x - 1)$.

Step4: Simplify the equation

Expand the right - hand side: $y - 1=-13x+13$. Then add 1 to both sides to get $y=-13x + 14$.

Answer:

C. $y=-13x + 14$