Proof Check: $4x^2+3x+17$ is continuous by using a $\epsilon$, $\delta$ Argument Can anyone double check my following epsilon delta proof.
I want to prove that the following function is continuous with an Epsilon Delta Argument.
$$ f: x \in \mathbb R \mapsto (4x^2+3x+17) \in \mathbb R $$
So I started with
$$\left\lvert f(x)-f(y) \right\rvert =$$
$$= \left\lvert (4x^2+3x+17) - (4y^2+3y+17) \right\rvert$$
$$= \left\lvert (4x^2+3x)-(4y^2+3y) \right\rvert$$
$$= 3\left\lvert (4x^2+x)-(4y^2+y) \right\rvert$$
$$= 3\left\lvert (\frac {4}{3}x^2+x)-(\frac {4}{3}4y^2+y) \right\rvert $$
$$= 3\left\lvert \frac {4}{3}x^2 - \frac {4}{3}y^2 + x-y \right\rvert$$
$$= 3\left\lvert \frac {4}{3}x^2 - \frac {4}{3}y^2 + x-y \right\rvert$$
(Triangel inequality)
$$\le 3 \left(\left\lvert \frac {4}{3}x^2 - \frac {4}{3}y^2 \right\rvert + \left\lvert x-y \right\rvert \right)$$
Note that $\delta \le 1$
$$\le 3 \left(\frac {4}{3}\left\lvert x^2 - y^2 \right\rvert + \delta \right)$$
$$= 3 \left(\frac {4}{3}\left\lvert (x - y) \right\rvert\left\lvert (x +y) \right\rvert + \delta \right)$$
$\delta \le 1$
$$\le 3 \left(\frac {4}{3}\left\lvert x + y \right\rvert \delta + \delta \right)$$
Adding zero in form of y - y
$$= 3 \left(\frac {4}{3}\left\lvert x + y - y + y \right\rvert \delta + \delta \right)$$
$$= 3 \left(\frac {4}{3}\left\lvert x - y + 2y \right\rvert \delta + \delta \right)$$
Triangel inequality
$$\le 3 \left(\frac {4}{3}(\left\lvert x - y \right\rvert +\left\lvert 2y \right\rvert) \delta + \delta \right)$$
Note that $\delta \le 1$
$$\le 3 \left(\frac {4}{3}(\delta +\left\lvert 2y \right\rvert) \delta + \delta \right)$$
$$= 4\delta^2+4\left\lvert 2y \right\rvert \delta + 3\delta $$
$$= \delta(4\delta+4\left\lvert 2y \right\rvert + 3) $$
Note that $\delta \le 1$
$$\le \delta(4 +4\left\lvert 2y \right\rvert + 3) $$
$$\le \delta(\left\lvert 8y \right\rvert + 7) = \epsilon$$
Therefore $$\delta = \frac {\epsilon}{(\left\lvert 8y \right\rvert + 7)} \;\; with \; \delta \le1$$
Thanks in advance for the help, I really appreciate it. :)
 A: Your solution looks good. The following steps could be simplified a bit to make it more readable.
\begin{align}
\left\lvert f(x)-f(y) \right\rvert
&=
3\left\lvert \frac {4}{3}x^2 - \frac {4}{3}y^2 + x-y \right\rvert
\quad (\textrm{why bother factoring out $3$?})\\
& \le 3 \left(\left\lvert \frac {4}{3}x^2 - \frac {4}{3}y^2 \right\rvert + \left\lvert x-y \right\rvert \right)\\
& \le 3 \left(\frac {4}{3}\left\lvert x^2 - y^2 \right\rvert + \delta \right)
\quad (\color{red}{\textrm{for $|x-y|<\delta$} })\\
&= 3 \left(\frac {4}{3}\left\lvert (x - y) \right\rvert\left\lvert (x +y) \right\rvert + \delta \right)\\
&= 3 \left(\frac {4}{3}\left\lvert x - y + 2y \right\rvert \delta + \delta \right)\\
&\le 3 \left(\frac {4}{3}(\left\lvert x - y \right\rvert +\left\lvert 2y \right\rvert) \delta + \delta \right)
\quad (\textrm{triangle inequality})\\
&\le 3 \left(\frac {4}{3}(\delta +\left\lvert 2y \right\rvert) \delta + \delta \right)\\
&=4(\delta +\left\lvert 2y \right\rvert) \delta + 3\delta
\\
&\le 
4(1+\left\lvert 2y \right\rvert) \delta + 3\delta
\quad (\delta\le 1)\\
&\le (7+|8y|)\delta
\end{align}
Set $$\delta =\min(\frac {\epsilon}{\left\lvert 8y \right\rvert + 7} ,1).$$
A: To prove that $f(x)=4 x^2 + 3 x + 17$ is continuous $\forall a\in\mathbb{R}$
$$\underset{x\to a}{\text{lim}}\left(4 x^2+3 x+17\right)=f(a)$$
Equivalently
$$\underset{h\to 0}{\text{lim}}\left(4 (a+h)^2+3 (a+h)+17\right)=4 a^2 + 3 a + 17$$
$\forall \varepsilon>0$ we must find a $\delta>0$ such that if $|h|<\delta$ then
$$|4 (a+h)^2+3 (a+h)+17-(4 a^2 + 3 a + 17)|<\varepsilon$$
that is
$$|8 a h+4 h^2+3 h|<\varepsilon$$
$$0<\delta<\left|\frac{1}{8} \sqrt{64 a^2+48 a+16 e+9}+\frac38-a\right|$$
if $|h|<\delta$ then $|f(a+h)-f(a)|<\varepsilon$
