What i did:
$f(x)=ax^2+bx+c$
$f(-2)=4a-2b+c=-10$
$f(0) =c > 0$
$f(1) =a+b+c > 0$
$f(2) =4a+2b+c > 0$
I thought using $b^2-4ac = 0$ for $f(-2)$ but its wrong since I am getting c = -6.
ANS: $-2x^2+4x+6$
What i did:
$f(x)=ax^2+bx+c$
$f(-2)=4a-2b+c=-10$
$f(0) =c > 0$
$f(1) =a+b+c > 0$
$f(2) =4a+2b+c > 0$
I thought using $b^2-4ac = 0$ for $f(-2)$ but its wrong since I am getting c = -6.
ANS: $-2x^2+4x+6$
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So $-1$ and $3$ are zeroes of this function so we can write it in factor form:
$$ f(x)=a(x+1)(x-3)$$ Now use $f(-2)=10$ to get $a$.