Question

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

Answer 1

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

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

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