The curve A (y = x3 – x2 + x -1) is perpendicular to the straight-line B at the point P (5, 2). If A and B intersect at P, what is the equation of B? Also, find any stationary points of the curve A.
Firstly find the gradient of A, through differentiation: dy/dy = 3x2 – 2x + 1. To find the gradient at P, substitute the x value of the P coordinate into this equation: dy/dx = 3(2)2 – 2(2) + 2 = 12 – 4 + 2 = 10. To find the perpendicular gradient, we must obtain the negative reciprocal, which in this case equals -1/10. As the equation of a straight line is y = mx + c, and we know m, we can set y = -x/10 + c, and to find c we substitute the values of a known point on line B, in this case the coordinates of P (5, 2): 5 = -2/10 + c, c = 26/5 = 5.2. Thus, the equation of line B is y = -x/10 + 26/5. Stationary points occur when the gradient equals zero, i.e. when dy/dx = 0. Therefore we shall set 3x2 – 2x + 1 = 0. To find the stationary points, we must find the x values using either the quadratic formula, or through factorising. Since the equation does not factorise, we will use the equation x = b +/- sqrt(b2 -4ac)/2a, using a = 3, b = -2 and c = 1. This gives us x = -2 +/- sqrt((-2)2-4(3x1))/2(3) which gives x = -2 +/- sqrt(-8)/6. Since the equation contains ‘sqrt(-8)’ the equation is invalid, and thus we can deduce that there are no stationary points. This can be checked on a graph plotter.
