What is the equation of the tangent at the point (2,1) of the curve with equation x^2 + 3x + 4.
dy/dx = 2x + 3, m = dy/dx = 7 Equation of tangent is a straight line hence we use y-y1 = m(x-x1), y-1 = 7(x - 2), y = 7x - 13
BA
dy/dx = 2x + 3, m = dy/dx = 7 Equation of tangent is a straight line hence we use y-y1 = m(x-x1), y-1 = 7(x - 2), y = 7x - 13
