Question

If $a<0$ and $b^2-4ac<0$, then the graph of $y=a{x}^2+bx+c$

• A. lies entirely below the $x$-axis
• B. lies entirely above the $x$-axis
• C. cut the $x$-axis
• D. touches the $x$-axis

Answer 1

Answer: (A)

lies entirely below the $x$-axis

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

Now $a{x}^2+bx+c=0$ implies

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Since $b^2-4ac<0$

Hence, there is no real value of $x$ for which $y=0$.

Therefore, the graph of $y=a{x}^2+bx+c$ never intersects the $x$-axis.

Now, $y=a{x}^2+bx+c$

$=a[x^2+\frac{b}{a}x+\frac{c}{a}]$

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

Since $b^2-4ac<0$, hence $4ac-b^2>0$

Also $(x+\frac{b}{2a}{)}^2>0$ (perfect square).

Hence whether $y$ is greater than or lesser than zero is dependent on the value of $a$.

Since it is given that $a<0$.

Hence, $a({(x+\frac{b}{2a})}^2+\frac{4ac-b^2}{4a^2})<0$

$\Rightarrow y<0$

Therefore, the graph of $y$ versus $x$ is always below $x$-axis.