Given \(\displaystyle y=px^2+qx-1\) and \(\displaystyle \frac{dy}{dx}= 7 at (1,3)\).FInd the value if \(\displaystyle p\) and \(\displaystyle q\).

Anyone help please!!

You know that \(\displaystyle (1, 3)\) lies on the curve \(\displaystyle y = p\,x^2 + q\,x - 1\).

So when you substitute the point, you find

\(\displaystyle 3 = p\,(1)^2 + q\,(1) - 1\)

\(\displaystyle 3 = p + q - 1\)

\(\displaystyle p + q = 4\).

You are also given information about the derivative - that the derivative is \(\displaystyle 7\) at the point \(\displaystyle (1, 3)\).

So \(\displaystyle y = p\,x^2 + q\,x - 1\)

\(\displaystyle \frac{dy}{dx} = 2p\,x + q\)

Now, substituting \(\displaystyle x = 1\) and \(\displaystyle \frac{dy}{dx} = 7\) we find

\(\displaystyle 7 = 2p(1) + q\)

\(\displaystyle 2p + q = 7\).

So now you have two equations in two unknowns, namely

\(\displaystyle p + q = 4\)

\(\displaystyle 2p + q = 7\).

Solve them simultaneously for \(\displaystyle p\) and \(\displaystyle q\).