How to find the interception points with $x$ axis of this function? There was a question in a certain exam to investigate the following function:
$$y=(x+2)\ln(x+2)-3x$$
But I stuck at finding the interception points with $y$ axis altough I checked with a function calculator and there are two although it doesn't tell what they are.
Maybe they're not supposed to be known just to somehow know the range where they should be but I don't know how to do that either.
The function according to symbolab:
https://www.symbolab.com/solver/functions-calculator/f%5Cleft(x%5Cright)%3D%5Cleft(x%2B2%5Cright)ln%5Cleft(x%2B2%5Cright)-3x/?origin=button
 A: Lambert W-Function
$$x=e^{W(-6/e^3)+3}-2=-2+e^{3+ProductLog[-6/e^3]}=-2+e^{3+\sum_{k=1}^{\infty}\frac{(-6)^k(-k)^{-1+k}}{e^{3 k}k!}}$$
$$x = e^{W_{-1}(-6/e^3)+3}-2=-2+e^{3+ProductLog[-1, -6/e^3]}$$
For example, you wanted to solve: $\ln(x)+ax=y$ add $\ln(a)$ to both sides to give:
$$\ln(ax)+ax=y+\ln(a)$$
Then exponentiate each side:
$$axe^{ax}=ae^y$$
So $ax=W(ae^y)$ and $x=\frac{1}{a}W(ae^y)$
A: If you do not want (or cannot) use Lambert function, consider the function $$y=(x+2)\log(x+2)-3x$$ its derivatives are $$y'=\log (x+2)-2\qquad , \qquad y''=\frac{1}{x+2}$$ The second derivative is always positive since, because of the logarithm $x>-2$. The first derivative cancels when $x=e^2-2$ and for this value $y=6-e^2$ which is negative. On the other side, if $x=1$, $y=3 \log (3)-3$ which is positive.
All of the above makes that there is one root $1<x_1<e^2-2$ and a second one $x_2>e^2-1$.
If, for simplicity, you compute the value of $y_k$ corresponding to $x_k=-2+e^k$, this gives $y_k=e^k (k-3)+6$ and, by inspection, you would notice that $y_2<0$ and $y_3>0$.
So, now, you are ready for solving for $x$ the equation. Newton method is probably the simplest to use. Starting from a "reasonable" guess $x_0$, the method will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$
So, for the first root, let us start using $x_0=1$; the method will then generate the following iterates $$x_1=1.32820$$ $$x_2=1.34993$$ $$x_3=1.35002$$ which is the solution for six significant figures.
For the second root, let us do the same using $x_0=e^3-2$; the iterates will then be $$x_1=12.0855$$ $$x_2=10.5328$$ $$x_3=10.3646$$  $$x_4=10.3623$$ which is the solution for six significant figures.
A: As many have noted, you may use the Lambert W-Function to find a "closed form" for the solution.
If you are confused on how this can have two roots that can be represented as a single equation, consider the quadratic formula, one formula for 2 solutions.
In fact, I might recommend the Lambert W-Function for this problem because it finds all of the zeroes, even the complex solutions.
For approximations, jimbo used one in his solution (the big summation thing).
But if you want easier to use solutions, just use something like linear approximation.  If you simply google search "linear approximation", you should find dozens of ways to do it.  Note that these are simply approximations and some may work better than others.
Also note that the only exact answer is in terms of the Lambert W-Function, but we use approximations for the Lambert W-Function's numerical answers (so whats really the point?).
