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[SOLVED] f(x)=px^2-qx find p and q


Well-known member
Jan 31, 2012
Given $f(x)=px^2-qx$, $p$ and $q$ are constants, point $A(1,3)$ lies on the curve, The tangent line to the curve at $A$ has gradient $8$. Find $p$ and $q$

well since it mentioned gradient then $f'(x)=2px-q$

then from $A(1,3)$ we have $3=p(1)^2-q(1)$ and from $m=8$, $8=2p(1)-q$

solving simultaneously we have

then $p=5$ and $q=2$

thus $f(x)=5x^2-2x$

View attachment 1097

no answer was given on this so just seeing if this is correct


"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
English is not my mother tongue so I'll just throw in my 2 cents. :)

Are you sure gradient means the slope of the tangent line? The interpretation I know of a gradient to a curve is the slope of the normal line to the curve. With this in mind we find that the tangent line has slope

$$m_{\parallel} = - \frac{1}{8}.$$

Therefore we arrive at the system

p-q = 3 \\
2p -q = - \frac{1}{8}.

Multiplying the bottom equation by 8 and subtracting the first from the second we get $16p - p + q - q = -1 -3$, thus $15 p = -4$ and $p = -4/15.$ Using the first equation we find

$$q = p - 3 = - \frac{4}{15} - 3 = - \frac{49}{15}.$$

It's a lot uglier than what you got, but it would be interesting to know whether this reading into the question is possible. :)



Well-known member
Jan 31, 2012
well you are probably correct...
I was assuming gradient and slope are the same thing

however. looks like the method is basically the same



Staff member
Feb 24, 2012
From what I've seen, gradient refers to the slope of the tangent line, not the normal to the slope of this line.