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Let $f:\mathbb R\to \mathbb R$ be a map defined as $f(x)=ax^2+bx+c$, we can rewrite this map as $f(x)=a(x+\frac{b}{2a})^2-\frac{\Delta}{4a}$, where $\Delta=b^2-4ac$. If $a\gt 0$, then the minimum value is took in $x= -\frac{b}{2a}$.

If $\frac{\Delta}{4a}$ were constant, then the question would be easy, because $a(x+\frac{b}{2a})^2\ge0$.

However, since this not the case, anyone can help me please?

Thanks a lot

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But $\Delta$ is constant and $a$ is constant. Why don't you think $\Delta/(4a)$ is constant? –  Antonio Vargas Jan 15 '13 at 9:16
    
@AntonioVargas you're right, of course. –  user42912 Jan 15 '13 at 9:34

2 Answers 2

up vote 2 down vote accepted

I add some information graphically just to complete the @Adi's theoretical approach.

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enter image description here

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More pictures! +1 –  amWhy Feb 17 '13 at 0:04

$$f(x)=ax^2+bx+c$$ $$f'(x)=2ax+b=0,x=-\frac{b}{2a}=0$$ then $$f(-\frac{b}{2a})=a\frac{b^2}{4a^2}-\frac{b^2}{2a}+c=\frac{4ac-b^2}{4a}$$ is extremum and if $f''(x)=2a>0,a>0$ that extrmum is minimum so $$f(x)>f\left(-\frac{b}{2a}\right),a>0$$

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Nice approach Adi. +1 –  Babak S. Jan 15 '13 at 10:01

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