Given $f(x)=ax^2+bx+c$ such that $\frac{df}{dx}=4$ at the point $(2,3)$ and $\frac{df}{dx}=-2$ at $(-1,0)$, find the constants $a, b$ and $c$.

Question

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Given that a function $f(x)=ax^2+bx+c$ is such that $\frac{df}{dx}=4$ at the point $(2,3)$ and $\frac{df}{dx}=-2$ at the point $(-1,0)$, find the value of the constants $a, b$ and $c$.

Answer 1

$$f(x) = ax^2 + bx+c$$ $$\frac{df}{dx} = 2ax+b$$ Substituting $\frac{df}{dx} = 4$ at $x=2$ $$2a(2) + b = 4$$ Substituting $\frac{df}{dx} = -2$ at $x=-1$ $$2a(-1) + b = -2$$ Solving the simultaneous equations, $a=1$, $b=0$, $f(x) = x^2 + c$ $$$$Substituting $(2,3)$: $$f(2) =2^2 + c = 3$$And so $c=-1$, making $f(x) = x^2-1$