Question

Let the graph of $f\left(x\right)=a{x}^2+bx+c$ passes through origin and makes an intercept of $10$ units on $x-$axis. If the maximum value of $f\left(x\right)$ is $25$, then the least value of $|a+b+c|$ is

Answer 1

$f\left(x\right)$ passes through origin
$\therefore f\left(0\right)=0\Rightarrow c=0$
$f\left(x\right)$ takes the maximum value
$\therefore a<0$
Now, $x-$ inetercept is $10$, so
The other root is either $10$ or $-10$.

$f\left(x\right)=kx(x+10)\text{or}kx(x-10)f\left(x\right)=k{x}^2+10kx\text{or}k{x}^2-10kx$
The maximum of the parabola is,
$-\frac{D}{4a}=25\Rightarrow \frac{4ac-b^2}{4a}=25\Rightarrow \frac{-100k^2}{4k}=25\Rightarrow k=-1$
Therefore,
$f\left(x\right)=-x^2-10x\text{or}-x^2+10x$
$a+b+c=-1+10+0\text{or}-1-10+0\to a+b+c=9,-11$

Hence, the least value of $|a+b+c|$ is $9.$