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 January 9th, 2014, 06:20 AM #1 Senior Member   Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 Gradient function. The gradient function of a curve is px^2-qx, where p and q are constants. The curve has a turning point at (2,-. The gradient of the tangent to the curve at the point where its x-coordinate =-1 is 8. Find the values of p and q , The equation of the curve
 January 9th, 2014, 12:51 PM #2 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Gradient function. That's pretty straight forward, isn't it? "The gradient of the function is $px^2- qx$".\ So the function itself is of the form $\frac{p}{3}x^3- \frac{q}{2}x^2+ c$ for constants p, q, and c. "The gradient of the tangent to the curve at the point where its x-coordinate =-1 is 8." Okay, the gradient is $px^2- qx$ so we must have $p(-1)^2- q(-1)= p+ q= 8$. "The curve has a turning point at (2,-." That tells us two things. First when x= 2, y= -8 so $\frac{p}{3}(2)^3- \frac{q}{2}(2^2)+ c= \frac{8p}{3}- 2q+ c= -8$. Also since it is a turning point, the derivative is 0: $p(2)^2- q(2)= 4p- 2q= 0$. Solve those three equations for p, q, and c. It would probably be easiest to solve p+ q= 8 and 4p- 2q= 0 for p and q first, then solve the third equation for c.

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