Find the equation of the tangent to the curve y = x^3 - 2x^2 + 3x - 1 at x = 1.
Correct: B
To find the equation of the tangent at x = 1, we first find the derivative dy/dx = 3x^2 - 4x + 3. At x = 1, the slope of the tangent is 3 - 4 + 3 = 2. The point on the curve at x = 1 is (1, 1). So, using point-slope form of a line, we have y - 1 = 2(x - 1), which gives y = 2x - 1.