Question

For any quadratic expression $y=a{x}^2+bx+c,a<0.$ It's range will be

• A. $[-\frac{D}{4a},\infty )$
• B. $(-\infty ,-\frac{D}{4a}]$
• C. $[\frac{D}{4a},\infty )$
• D. $(-\infty ,\frac{D}{4a}]$

Answer 1

The correct option is B $(-\infty ,-\frac{D}{4a}]$

Given : $y=a{x}^2+bx+c,a<0$

$y=a(x^2+\frac{b}{a}x+\frac{c}{a})$

$y=a(x^2+2.\frac{b}{2a}x+\frac{b^2}{4a^2}-\frac{b^2}{4a^2}+\frac{c}{a})$

$y=a({(x+\frac{b}{2a})}^2-\frac{b^2}{4a^2}+\frac{4ac}{4a^2})$

$y=a({(x+\frac{b}{2a})}^2-(\frac{b^2}{4a^2}-\frac{4ac}{4a^2}))$

$y=a({(x+\frac{b}{2a})}^2-\left(\frac{b^2-4ac}{4a^2}\right))$

$y=a({(x+\frac{b}{2a})}^2-\left(\frac{D}{4a^2}\right))$
Therefore the maximum value occurs when
$x+\frac{b}{2a}=0$
$\Rightarrow x=-\frac{b}{2a}$
and at this point $y_{max}$ is $-\frac{D}{4a}$
And $y\to -\infty$ as $x\to \pm \infty$

$\therefore Range\in (-\infty ,-\frac{D}{4a}]$