Given $$f(x)=ax^2+bx+c\ ; \quad a\neq0.$$ Prove that it is bijective if $$x \in \Bigg[\frac{-b}{2a},\ \infty \Bigg]$$ and $$ranf=\Bigg[\frac{4ac-b^2}{4a},\ \infty \Bigg).$$

I can prove that the range of $f(x)=ax^2+bx+c$ is $ranf=\Big[\frac{4ac-b^2}{4a},\ \infty \Big)$, if $a\neq0$ and $a\gt0$ by completing the square, so I know here that the leading coefficient of the given function is positive.

I have also proved that $f(x)=ax^2+bx+c$ is injective where $f:\big[0, \infty \big)\to\Bbb R.$

But I don't know how to prove that the given function is surjective, to prove that it is also bijective.

I admit that I really don't know much in this topic and that's why I'm seeking help here. An advanced thanks to those who'll take time to help me. :)

  • $\begingroup$ A quadratic is never surjective. Also: it is not true that $ax^2+bx+c$ is injective for all choices of $a,b,c$, even if you restrict your domain to $x>0$. For example, $f(x)=(x-1)(x-2)$. I think your difficulties stem from the fact that you have no picture of a quadratic in front of your eyes. $\endgroup$ Aug 9, 2018 at 14:10
  • $\begingroup$ But is the given quadratic in the question bijective when it has $x \in \Big[\frac{-b}{2a}, \infty \Big)$ and its range is $\Big[\frac{4ac-b^2}{4a}, \infty \Big)$ ? $\endgroup$
    – ShiroKuro
    Aug 9, 2018 at 14:23

1 Answer 1


As uniquesolution pointed out in the comments, a quadratic function cannot be surjective onto $\mathbb R$ (think of a picture of a parabola: it never reaches the $y$-values below/above its vertex). But it can be surjective onto $\left[\frac{4ac-b^2}{4a},\infty\right)$, which you seem to have already shown if you have shown that is indeed the range.

  • $\begingroup$ So, does it mean that once I have proved that $\Big[\frac{4ac-b^2}{4a}, \infty \Big)$ is indeed the range of the given quadratic, I also have proved that it is surjective? I apologize for not getting the point easily $\endgroup$
    – ShiroKuro
    Aug 9, 2018 at 14:20
  • $\begingroup$ @ShiroKuro For a function to be surjective onto a set $B$ means that the range of the function is the entire set $B$. If you have proved that the range of the function is the entire set $\left[\frac{4ac-b^2}{4a},\infty\right)$, then the function is surjective onto that set (but not surjective onto $\mathbb R$). Does this clarify the point? $\endgroup$
    – BallBoy
    Aug 9, 2018 at 14:22
  • $\begingroup$ Oh.. Yes! It does indeed! Thank you so much! But may I ask how I can prove that it is also injective? So I can finally prove that the given quadratic is a bijective function. $\endgroup$
    – ShiroKuro
    Aug 9, 2018 at 14:26
  • $\begingroup$ @ShiroKuro What was your original attempt to prove injectivity? $\endgroup$
    – BallBoy
    Aug 9, 2018 at 14:28
  • $\begingroup$ I tried to prove that $f(x_1)=f(x_2)$ where $ax_1^2+bx_1+c=ax_2^2+bx_2+c.$ But I always get tangled up somewhere along the process, and have gotten almost nowhere. $\endgroup$
    – ShiroKuro
    Aug 9, 2018 at 14:31

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