How does this proof work that shows a function is onto? > Is the function $p:\mathbb{R} \to \mathbb{R}$ defined by $p(x) = 3x^2 - 4x + 5$ onto? > > Let $y = 3x^2 - 4x + 5$, we want to know if we can always express $x$ in terms of $y$. Rearranging the equation, we find > > $$ 3x^2 - 4x + (5 - y) = 0 \tag{1} $$ > > We want this equation to be solvable over $\mathbb{R}$, that is, we want its solutions to be real. This requires its discriminant to be nonnegative. So we need: > > $$ (-4)^2 - 4 \cdot 3 \cdot (5 - y) = 12y - 44 \ge 0 \tag{2} $$ > > We have solutions only when $y \ge \frac{11}{3}$. This means, when $y \lt \frac{11}{3}$ we cannot find an $x$-value such that $p(x) = y$. Therefore, $p$ is not onto. I understand that the rearranged equation in $(1)$ is a quadratic, so we need to have its discriminator non-negative in order to have real solutions. But what's happening in $(2)$ and how do they then solve for $\frac{11}{3}$, this part I don't get.

The proof begins by attempting to find the pre-image of $p(x)$, i.e. given any $y \in \mathbb R$, you want the set of $x$ such that $p(x)=y$. If this set is never empty, then $p(x)$ is onto.

So solving (1) gives the required set of $x$'s. But to know whether any solutions exist in $\mathbb{R}$, we need to check that the discriminant is non-negative. This is done in (2), and for the discriminant to be non-negative, (after some algebra) we ultimately need to satisfy the inequality $12y-44 \geq 0$.

However, this inequality can't be satisfied if $y<\frac{11}{3}$, i.e. the there are no (real) values of $x$ that give $p(x)=y$ if you choose $y$ to be any value less than $\frac{11}{3}$. In other words, $p(x)$ only maps onto $[\frac{11}{3},\infty)$, and not the entire real line.