The curve C has equation y = 2x^2 - 12x + 16 Find the gradient of the curve at the point P (5, 6).
The most simple way to find the gradient of a quadratic curve at a point is to differentiate. Through differentiation a function for the gradient at each value of x can be found. 'y' must be isolated on one side of the equation for the differentiation to be simple, as is the case for the equation we are given here for curve C. Through differentiation this 'y' becomes 'dy/dx', a way of representing the gradient (change in y/ change in x) at different points. Then the right side of the equation must be differentiated by multiplying each x value by its 'power' and then taking away one from each x value's power.For curve C, the 2x2 becomes 4x (as 22=4 and 2-1=1), the -12x becomes -12 (as 121=12 and 1-1=0) and the 16 becomes 0 (as this is 16x0 and 16*0=0). This gives the differentiated function dy/dx = 4x - 12. To then find the gradient at point P, the value for x at point P simply needs to be substituted in to the differentiated function. This gives dy/dx = 4(5) - 12, and solving this gives dy/dx = 20 - 12, dy/dx = 8. Therefore the gradient of the curve at point P is equal to 8.
