# Differentiation problem

#### mastermin346

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

#### Prove It

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

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 $$\displaystyle q$$.

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