Question

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

Answer 1

\(\because\) 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 \(-\dfrac b{2a}>0\Rightarrow b<0\)

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