Find the stationary points on the curve y = x^3 + 3x^2 - 9x - 4

A stationary point is where the gradient is exactly zero - the curve is neither increasing or decreasing. This means that we need to differentiate y to find dy/dx and then set this equal to 0. Doing this, using the normal rules for differentiation, we would get 3x^2 + 6x -9 Then, we would set this equal to zero and factorise the equation to find out the x values of our stationary points. Doing this, we get (x + 3)(x - 1) = 0 Leaving x=-3 or x=1 Finally, substitute these values into y = x^3 + 3x^2 -9x -4 and this will give you the y-coordinates to the stationary points. The final answers are therefore (-3,23) and (1,-9)

RR

Related Maths A Level answers

All answers ▸

integrate e^x sin x dx


How do I sketch the graph y = (x^2 + 4*x + 2)/(3*x + 1)


Find the angle between two lines with vector equations r1 = (2i+j+k)+t(3i-5j-k) and r2 = (7i+4j+k)+s(2i+j-9k)


how do you differentiate y=x^2 from first principles?