Question

The graph of the quadratic polynomial $y=a{x}^2+bx+c$ has its vertex at $(4,-5)$ and two $x$ intercepts, one positive and one negative. Which of the following hold(s) good?

• A. $a>0$
• B. $bc>0$
• C. $2b+c+5=0$
• D. $b+8a=0$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $a>0$
B $bc>0$
C $2b+c+5=0$
D $b+8a=0$

$\because$ Vertex is in $4\text{th}$ quadrant and it has two real roots also.
$\therefore a>0$
So, the approximate graph of the polynomial will be


From the graph $y$ intersection is $-ve$
$\therefore c<0$
and $-\frac{b}{2a}>0\Rightarrow b<0$

Coordinates of the vertex is $(-\frac{b}{2a},-\frac{D}{4a})\equiv (4,-5)$
$\Rightarrow \frac{b}{2a}=-4,\frac{b^2-4ac}{4a}=5\Rightarrow b+8a=0,\frac{b^2}{4a}=5+c\Rightarrow b+8a=0,2b+c+5=0$