The curve y = 2x^3 -ax^2 +8x+2 passes through the point B where x = 4. Given that B is a stationary point of the curve, find the value of the constant a.
y=2x3 - ax2 + 8x +2. If we are told that B is the stationary point of the curve, then it is at that point where the gradient of the curve is equal to 0. In order to find the gradient of the curve at this point we must differentiate. Thus we have the equation dy/dx = 6x2 - 2ax + 8. In order to get that equation you must times the integer in front of the x by the numerical value of the power. Then after doing so, you reduce the power by one.
We know that when x=4, the gradient is equal to 0. Therefore, you simply substitute the 4 in the equation and you get 96 - 8a + 8 = 0. Then you want to get your known values on one side and the unknown on the other.Thus you get 104 = 8a104/8 = aa= 13
