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Let $f(x)=ax^2+bx+c$ where $a \neq 0$. What is the range of $f$?

I know that the answer is if $a<0$ then the range is $\left(-\infty,\frac{4ac-b^2}{4a}\right]$ and if $a>0$ the range is $\left[\frac{4ac-b^2}{4a},\infty\right)$ by graphing or the fact that the function has constant second derivative.

I would like to find the range without using any calculus or results from calculus.

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    How do you want to find the range without using any calculus? In fact this function is not monotone. – 2017-01-22
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    @BehrouzMaleki Ah shit, I meant constant second derivative. Let me edit. I have no idea, thats why I'm asking. – 2017-01-22

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HINT: use complete square: $$a\left(x^2+2\frac{b}{2a}x+\frac{c}{a}\right)=a\left(\left(x+\frac{b}{2a}\right)^2+\frac{c}{a}-\frac{b^2}{4a^2}\right)=a\left(x+\frac{b}{2a}\right)^2+\frac{4ac-b^2}{4a}$$ i think this should help

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    I noticed this but I didn't quite see how this would help finish the argument. – 2017-01-22
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    you can use your arguuments above to finish the proof – 2017-01-22
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    Ahhh I see it actually. If $a>0$ I can show that $\frac{4ac-b^2}{4a}-1$ is not in the image of the function and similarly for $a<0$. Thanks! – 2017-01-22