The curve C has equation y=2x^2 -11x +13. (a) The point P has coordinates (2, – 1) and lies on C. Find the equation of the tangent to C at P.
We must firstly understand that the tangent at point P is linear, and we must use y = mx +c format for the equation. To calculate the equation we need to find the unknowns m and c. To find the unknown m which represents the gradient, we must differentiate equation (a), y=2x2 -11x+13; dy/dx = 4x-11 We know where we would like to find the gradient at point P, which has a corresponding x value of 2. We may input our value for x=2 into dy/dx = 4x-11; dy/dx = 4(2)-11; dy/dx=-3 i.e. our gradient, m at point P. To find the intercept c, we must input a co-ordinate we know on the tangent, co-ordinate P, and our gradient, recently calculated using differentiation as m = -3. Inputting gives us, y = mx + c ; (-1) = (-3)(2) + c ; c = 5 Therefore our equation of the tangent at point C is: y = -3x +5.
