Question

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

• A. $a>0$
• B. $b<0$
• C. $c<0$
• D. $8a=b$

Answer 1

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

The correct options are
A $a>0$
B $b<0$
C $c<0$

Since the graph of the quadratic trinomial $y=a{x}^2+bx+c$ has two x-intercepts, one positive and one negative, So $y<0$ at $x=0$
it implies that $c<0$

Since its vertex is at $(4,-5)$ ,so graph is upward parabola i.e. $a>0$
according to properties of the graph of quadratic equation its position $x$ of vertex is $\frac{-b}{2a}$
since, given that position $x$ of vertex is $4$. i.e. positive.
So, $\frac{-b}{2a}>0$

Now. $a>0$ so $b<0$.
$\frac{-b}{2a}=4$
$\Rightarrow$ $b=-8a$

Options A,B,C are correct.