Let f(x)=ax^2+bx+c be such that f(1)=3, f(-2)= and f(3)=4.

Question

Let $f(x)=ax^2+bx+c$ be such that $f(1)=3,\ f(-2)=\lambda$ and $f(3)=4$. If $f(0)+f(1)+f(-2)+f(3)=14$, then $\lambda$ is equal to:

Accepted Answer

$f(1)=a+b+c=3$; $f(3)=9a+3b+c=4$.

$f(0)+f(1)+f(-2)+f(3)=14 \implies c+3+(4a-2b+c)+4=14 \implies 4a-2b+2c=7$.

From $f(1)$ and $f(3)$: $8a+2b=1$. Combined with $4a-2b+2c=7-2(a+b+c-3)$... solving the system gives $a=\dfrac16,\ b=-\dfrac16,\ c=3$.

$$f(-2)=4a-2b+c=\frac46+\frac26+3=1+3=4.$$

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