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Polynomial Differentiation: Gradient and Normal
The x-values of the points on the curve y = 3 x3 + 3 x2 + 4 x – 8 where the gradient is 8 are
x = and x = to 3 dps.

tickcross



Worked Solution

Differentiating gives a formula for the gradient at point (x, y):
dy
---
dx
 = 9 x2 + 6 x + 4

So we need to solve 9 x2 + 6 x + 4 = 8. Using the Quadratic Formula:
If a x2 + b x + c = 0 then x =
–b ± √(b2 – 4 ac)
---
2 a
Here a = 9, b = 6 and c = 4 – 8 = –4 and the formula gives the roots:
x = 0.412 and x = –1.079 to 3 dps.

© MEI Produced by Dr Ron Knott, revised 3 April 2007
tom . button [AT] mei . org . uk 		Test reference: PolyGrads.html?ref=4639