Suppose a parabola y = a x 2 + b x + c has two x intercepts, one positive and one negative, and its vertex…

Given, $y=a{x}^2+bx+c$ has two $x$-intercepts one is positive and one is negative and vertex is $(2,-2)$. The graph of $y=a{x}^2+bx+c$ is

Clearly from graph

$c>0$

$-\frac{b}{a}>0\Rightarrow -b>0$

$\Rightarrow b<0$

$f\left(1\right)=a+b+c<0$

$ab<0,ac<0,bc>0$