Find the equation of the normal to the curve y = 4x3 − 5x2 + x at the point (1, 2).

Given the curve:

$$y = 4x^3 - 5x^2 + x$$

Differentiate $y$ with respect to $x$:

$$\frac{dy}{dx} = \frac{d}{dx}(4x^3) - \frac{d}{dx}(5x^2) + \frac{d}{dx}(x)$$ $$\frac{dy}{dx} = 12x^2 - 10x + 1$$

Substitute $x=1$ into the derivative:

$$m_{tangent} = 12(1)^2 - 10(1) + 1$$ $$m_{tangent} = 12 - 10 + 1$$ $$m_{tangent} = 3$$

(You can use an fx-991 calculator to find the derivative at $x=1$ by inputting $\frac{d}{dx}(4x^3 - 5x^2 + x)|_{x=1}$, which will give 3.)

The normal line is perpendicular to the tangent line. The product of their slopes is $-1$.

$$m_{normal} = -\frac{1}{m_{tangent}}$$ $$m_{normal} = -\frac{1}{3}$$

The point is $(x_1, y_1) = (1, 2)$ and the slope is $m_{normal} = -\frac{1}{3}$.
The point-slope form is $y - y_1 = m(x - x_1)$.

$$y - 2 = -\frac{1}{3}(x - 1)$$

Multiply both sides by 3 to eliminate the fraction:

$$3(y - 2) = -1(x - 1)$$ $$3y - 6 = -x + 1$$

Add 6 to both sides:

$$3y = -x + 1 + 6$$ $$3y = -x + 7$$

This matches Option A.