To show $ax^{2}+bx+c$ is neither injective or surjective. I want to show that $\displaystyle ax^{2}+bx+c$ is nither injective nor surjective where $a,b,c\in \mathbb{R}$ and $a\neq 0$. Can i use the following result? > If $f$ is a continuous real valued function on interval $I$ and if $f'(x)>0$ for all $x$ in $I$ except possibly at the end points of $I$, then $f$ is one-to-one. Here $f(x)=ax^{2}+bx+c$ being a polynomial so it is continuous in $R$ but $f^{'}(x)=2ax+b$ attains both signs that's why $f$ is not injective. And for surjective: for $y=ax^{2}+bx+c $, there does not exist any $x\in R$ such that $f(x)=y.$ Am i right?

If you complete the square, you will know all. We have $$ax^2 + bx + c = a\left(x^2 + {b\over a} x \right) + c = a\left(x + {b\over 2a}\right)^2 + c - {b^2\over 4a}. $$ If $f(x) = ax^2 + bx + c$, we have $$f(x) = a\left(x + {b\over 2a}\right)^2 + {4ac - b^2\over 4a}.$$ if $a > 0$ then $f(x)\ge {4ac-b^2\over 4a}.$ This precludes $f$ from being onto. If $a < 0$ a similar phenomenon inheres. So, $f$ is never onto.

Choose any two points equidistant from $-{b\over 2a}$. Then $f$ takes them to the same value. Therefore $f$ is never 1-1.