Question

If the curve $y=a{x}^2+bx+c=0$ has $y$-intercept $6$ and vertex as $(\frac{5}{2},\frac{49}{4}),$ then the value of $a+b+c$ is

Answer 1

$y=a{x}^2+bx+c=0$
$y$-intercept $=6$
$\Rightarrow c=6$
$(-\frac{b}{2a},-\frac{D}{4a})=(\frac{5}{2},\frac{49}{4})$
$\Rightarrow -\frac{b}{2a}=\frac{5}{2}\Rightarrow b=-5a$
and $-\frac{(b^2-4ac)}{4a}=\frac{49}{4}$
$\Rightarrow c-\frac{b^2}{4a}=\frac{49}{4}$
$\Rightarrow 6-\frac{25a^2}{4a}=\frac{49}{4}(\because b=-5a)$
$\Rightarrow a=-1$
$\Rightarrow b=5$
$\therefore y=-x^2+5x+6$
$\Rightarrow a+b+c=10$