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The problem goes like this:

Let $a, b$ and $c$ be given numbers, where $a>0$, and let $x$ be some number. What is the smallest possible value of $ax^2+bx+c$ ?

The terms 'given number' and 'some number' really bother me. Now I'm just really confused about this. Any help would be appreciated, thanks!

Brian Tung
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Kurns
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3 Answers3

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Let $f(x) = ax^2+bx+c = a\left(x+\dfrac{b}{2a}\right)^2 - \dfrac{b^2-4ac}{4a}$. Can you conclude from this?

DeepSea
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Since $a>0$, the parabola opens up. That means the vertex of the parabola is the minimum of the function. Can you find the vertex of $f(x)$ in terms of $a, b,$ and $c$?

scott
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Assuming the title question should refer to $ax^2+bx+c$:

That expression describes a concave-upward parabola (an upright U-shape, in other words). Its lowest point is therefore at the vertex of the parabola; you may recall that its $x$-coordinate is $-\frac{b}{2a}$ (which accounts for that fraction's appearance in the quadratic formula), and therefore the $y$-coordinate is what comes out when you plug $x = -\frac{b}{2a}$ into $ax^2+bx+c$.

For instance, if we consider the specific case $x^2-4x+9$, the vertex has an $x$-coordinate of $-\frac{-4}{2\cdot1} = 2$, and its $y$-coordinate is therefore $(2)^2-4(2)+9 = 5$. Now you just need to generalize for $a, b, c$.

Brian Tung
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  • That helped, but as English isn't my native language and this field of mathematics is not my strongest ground, could you please take the answer a little further? – Kurns Jun 09 '16 at 00:31
  • How do I exactly 'generalize'? – Kurns Jun 09 '16 at 00:53