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?
 A: 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.
A: You could try a naive approach. Put $\,f(x):=ax^2+bx+c\,\,,\,a\neq 0\,$ , so that
$$f(p)=f(q)\Longleftrightarrow ap^2+bp=aq^2+bq\Longleftrightarrow (p-q)[a(p+q)+b]=0 \Longrightarrow$$
$$\Longrightarrow p=q\,\,\,\,\,\,\text{or}\,\,\,\,p+q=-\frac{b}{a}$$
Since $\,a\neq 0\,$ the second option above is always possible and thus the function cannot be $\,1-1\,$ .
Finally, it's an easy exercise in geometry (not even basic calculs needed!) to show that 
$$ \left(-\frac{b}{2a}\,,\,-\frac{\Delta}{4a}\right)\,\,,\,\,\Delta:=b^2-4ac$$
is a maximum point or a minimum point, according as whether $\,a<0\,\,\,or\,\,\,a>0\,$ , resp., thus rendering $\,f(x)\,$ not onto.
A: Hint $\ $ There are bijections $\rm\:g,h\:$ such that $\rm\:h(f(g(x)) = x^2,\:$ so if $\rm\:f\:$ is in/sur-jective so too is $\rm\:x^2\:$ since in/sur-jective maps are closed under composition.
A: The theorem you posted is not actually useful because we want to show the opposite direction. However, looking at the limits is quite useful and even more fundamental than using differentiability.
Let's take a look at the case $a>0$. Then, all you need to know is that $f(x) \to +\infty$ as $x \to \pm \infty$ and that $f$ is continuous. It follows that $f$ has some global minimum $y$ (how?) 


*

*Can $f$ be injective? No, because it approaches $+\infty$ in both directions, so it passes through every value in $[y,\infty)$ at least twice. (why?)

*Can $f$ be surjective? No, because it never attains any value below $y$.
A: \begin{eqnarray}
f(x)= ax^2+bx+c&=& a[x^2+\dfrac{b}{2a}x+ \dfrac{c}{a}] \\
&=& a[(x+\dfrac{b}{2a})^2 -\dfrac{b^2-4ac}{4a}] \\
&\ge& \dfrac{b^2-4ac}{4}
\end{eqnarray}
Hence, $f$ is not injective and as $f(x)=f(-x-2\dfrac{b}{2a}), f$  is not surjective.
A: Consider the graph $y=ax^2+bx+c$ of your function, and find a change of coordinates
$$
\begin{cases}
X=x-\alpha \\
Y=y-\beta
\end{cases}
$$
such that $y=ax^2+bx+c$ becomes $Y=aX^2$. This is clear if you know that your curve is a parabola, so that you just need to translate its vertex into the origin: $\alpha$ and $\beta$ are related to the vertex coordinates.
Now, it is evident that $Y=aX^2$ describes a function which is neither injective, nor surjective: $Y\geq 0$, so that the range cannot contain negative values. And $Y(-1)=Y(1)$, for instance.
Reversing the change of coordinates, you are done. 
