The graph of the quadratic polynomial y=ax2+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
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B
bc>0
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C
2b+c+5=0
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D
b+8a=0
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Solution
The correct option is Db+8a=0 ∵ Vertex is in 4th quadrant and it has two real roots also. ∴a>0
So, the approximate graph of the polynomial will be
From the graph y intersection is −ve ∴c<0
and −b2a>0⇒b<0
Coordinates of the vertex is (−b2a,−D4a)≡(4,−5) ⇒b2a=−4,b2−4ac4a=5⇒b+8a=0,b24a=5+c⇒b+8a=0,2b+c+5=0