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Let $a$, $b$ and $c$ be real numbers and consider that $f$ maps $\mathbb{R}$ to $\mathbb{R}$.

For what values of $a$, $b$ and $c$ is $f(x) = ax^2 + bx + c$ (i) one-to-one? (ii) onto?

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It would be great if you could show us some of the work you've done. Have you considered what shape the graphs of the functions take for different values of $a,\ b$ and $c$? –  EuYu Nov 19 '12 at 5:20
    
I was thinking if a = 0 and b is not equal to zero, that is one case when f is both one-to-one and onto, but i couldn't figure out how to formulate the argument when a is not equal to zero. Also i tried to do an algebraic solution but ran out of ideas? –  UH1 Nov 19 '12 at 5:30
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1 Answer

Informal, to give you a start:

If $a\ne 0$, what does the curve $y=ax^2+bx+c$ look like? Bad, no? It is an upward or downward opening parabola, and one can see that both one to one and onto fail.

So for one to one, or onto, we need $a=0$. Suppose from now on that $a=0$.

If $b=0$, big trouble.

Show that if $a=0$ and $b\ne 0$, the function is one to one and onto. Geometrically, $y=bx+c$ is a line neither up and down nor parallel to the $x$-axis. The value of $c$ doesn't matter.

After you have figured out geometrically what's going on, doing the algebraic details (if required) will not be difficult.

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thanks but as i commented above, i already knew the part when a is equal to zero. But i was not sure when a is not equal to zero but i think i agree with your claim that if a is not equal to zero then there is no chance of it being one-to-one or onto. right? –  UH1 Nov 19 '12 at 5:49
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If you already knew the part when $a=0$, the time and place to acknowledge this was when you posted the question, not after someone took the time to think it through and type it out. –  Gerry Myerson Nov 19 '12 at 5:51
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@UH1: If you want algebra, suppose $a\ne 0$, and set $ax^2+bx+c=y$, or equivalently $ax^2+bx+c-y=0$. Solve using the Quadratic Formula. We get something that involves $\pm\sqrt{b^2-4a(c-y)}$. Argue that there are $y$ such that the thing under the square root sign is negative, so no (real) $x$. That shows not onto. Argue also that there is a $y$ such that the thing under the square root sign is positive. Then $2$ solutions, therefore not one to one. –  André Nicolas Nov 19 '12 at 5:57
    
@AndréNicolas: Thank you so much, appreciate your help a lot. –  UH1 Nov 19 '12 at 16:42
    
And @Gerry Myerson: You are right and I apologize for this, its just that I am new at using this forum, won't happen again. Thanks for pointing it out. –  UH1 Nov 19 '12 at 17:38
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