What is the equation of the line that is tangent to the curve y = x^2 at the point (1, 1)?
Correct: B
To find the equation of the line that is tangent to the curve y = x^2 at the point (1, 1), we need to find the slope of the tangent line. The slope of the tangent line is given by the derivative of the curve, which is dy/dx = d(x^2)/dx = 2x. Evaluating this derivative at x = 1, we get a slope of 2. The equation of a line with slope m and passing through the point (x1, y1) is given by y - y1 = m(x - x1). Plugging in the slope and the point (1, 1), we get y - 1 = 2(x - 1), which simplifies to y = 2x - 1.