Question

The graph of $f\left(x\right)=a{x}^2+bx+c$ is shown below, such that $b^2-4ac=-4.$ If the length of segment $AB$ and $AC$ are $1$ and $4$ respectively, then the value of $(a+b+c)$ is equal to

• A. 26.00
• B. 26.0
• C. 26

Answer 1

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

$f\left(x\right)=a{x}^2+bx+c$
$D=b^2-4ac=-4...\left(1\right)$
Least value of $f\left(x\right)=-\frac{D}{4a}=\overline{AB}$
$\Rightarrow -\frac{D}{4a}=1$
$\Rightarrow -(-4)=-4a\left[\text{From}\right(1\left)\right]$
$\Rightarrow a=1$

Also $-\frac{b}{2a}=-4\Rightarrow b=8$
And $b^2-4ac=-4$
$\Rightarrow 64-4c=-4$
$\Rightarrow c=17$
Hence, $a+b+c=1+8+17=26$