Question

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

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

Answer 1

Answer: (B)

touches the $x$-axis and lies above it

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 single (equal roots) root to the above equation.

Therefore, the graph of $y=a{x}^2+bx+c$ intersects the $x$ axis at only one point.

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 above $x$ axis.