Solve for 42 - 50

Given the curve $y = x^3 - 2x^2 + 3x - 1$.
Differentiate $y$ with respect to $x$:

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

Substitute $x=2$ into the derivative:

$$m = 3(2)^2 - 4(2) + 3$$ $$m = 3(4) - 8 + 3$$ $$m = 12 - 8 + 3$$ $$m = 7$$

The point-slope form is $y - y_1 = m(x - x_1)$.
Using the point $(x_1, y_1) = (2, 5)$ and the slope $m=7$:

$$y - 5 = 7(x - 2)$$ $$y - 5 = 7x - 14$$ $$y = 7x - 14 + 5$$ $$y = 7x - 9$$

The equation of the tangent to the curve at the point $(2, 5)$ is: