Question

Suppose a parabola $y=a{x}^2+bx+c$ has two x intercepts one positive and one negative, and its vertex is $(2,2)$. Then which of the following is true?

• A. $ab>0$
• B. $bc>0$
• C. $ca>0$
• D. $a+b+c>0$

Answer 1

Answer: (B)

$bc>0$

Vertex is at $(2,2)$ and the parabola cuts the $x$ axis so its is a downward opening parabola .

For a downward opening parabola $a<0$

At $(2,2)$ $\frac{dy}{dx}=0$

$\Rightarrow 2ax+b=04a+b=0b=-4aa<0\Rightarrow b>0$

For such a parabola you can clearly see from the figure the $y$ intercept can only be positive.

So $c>0$

Now we have $a<0,b>0$ and $c>0$

$\Rightarrow bc>0$

So option $\left(B\right)$ is correct