what is the equation of the normal line to the curve y=x^2-4x+3 at the point (5,8)?

first we need to work out the gradient of the tangent at (5,8). so first differentiate the equation once with respect to x. this gives is dy/dx=2x-4. so at (5,8) the gradient of the tangent is 6. this means that the gradient of the normal is -1/6 as the normal line and the tangential line is perpendicular and thus there gradients are negative reciprocals of each other. thus the equation of the normal is of the form y=-1/6x+c where c is a constant to be found. to get this constant we substitute in the point (5,80. thus 8=-1/6(5)+c. so c=56/6 (or 8.33333333...). so the equation is y=-1/6c+53/6 or can be written as 6y=-x+53.