# [SOLVED]f(x)=px^2-qx find p and q

#### karush

##### Well-known member
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
$3=p-q$
$8=2p-q$

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

#### Fantini

MHB Math Helper
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

$$\begin{cases} p-q = 3 \\ 2p -q = - \frac{1}{8}. \end{cases}$$

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.

Cheers!

#### karush

##### Well-known member
well you are probably correct...
I was assuming gradient and slope are the same thing

however. looks like the method is basically the same

K