Quadratic Function Real Life Examples and Applications

Quadratic Variation in SDEs.
Let $X_t$ be a stochastic process satisfying:
${\displaystyle X_t=X_0+{\int <!--\; \int \; -->}_0^t\mathrm{\mu <!--\; \mu \; -->}(s,\mathrm{\omega <!--\; \omega \; -->})\mathrm{d}s+{\int <!--\; \int \; -->}_0^t\mathrm{\nu <!--\; \nu \; -->}(s,\mathrm{\omega <!--\; \omega \; -->})\mathrm{d}{B}_s}$
Shorthand: $\mathrm{d}{X}_t={\mathrm{\mu <!--\; \mu \; -->}}_t\mathrm{d}t+{\mathrm{\nu <!--\; \nu \; -->}}_t\mathrm{d}{B}_t$
where the last integral is a Brownian motion integral, and where $\mathrm{\mu <!--\; \mu \; -->}(t,\mathrm{\omega <!--\; \omega \; -->}),\mathrm{\nu <!--\; \nu \; -->}(t,\mathrm{\omega <!--\; \omega \; -->})$ are ${\mathcal{F}}_t$ adapted $L^2$ functions. (In class we so far only did the case where $\mathrm{\mu <!--\; \mu \; -->}$ and $\mathrm{\nu <!--\; \nu \; -->}$ are deterministic).
Let f(t,x) be a twice differentiable deterministic function.
The usual presentation of Ito's lemma is:
${\displaystyle \mathrm{d}f(t,X_t)=\left(\frac{\mathrm{\partial <!--\; \partial \; -->}f}{\mathrm{\partial <!--\; \partial \; -->}t}+{\mathrm{\mu <!--\; \mu \; -->}}_t\frac{\mathrm{\partial <!--\; \partial \; -->}f}{\mathrm{\partial <!--\; \partial \; -->}x}+\frac{1}{2}\frac{{\mathrm{\partial <!--\; \partial \; -->}}^{2}f}{\mathrm{\partial <!--\; \partial \; -->}{x}^2}{\mathrm{\nu <!--\; \nu \; -->}}_t^2\right)\mathrm{d}t+\frac{\mathrm{\partial <!--\; \partial \; -->}f}{\mathrm{\partial <!--\; \partial \; -->}x}{\mathrm{\nu <!--\; \nu \; -->}}_t\mathrm{d}{B}_t}$
The professor offered us this shorthand:
${\displaystyle \mathrm{d}f(t,X_t)=\frac{\mathrm{\partial <!--\; \partial \; -->}f}{\mathrm{\partial <!--\; \partial \; -->}t}\mathrm{d}t+\frac{\mathrm{\partial <!--\; \partial \; -->}f}{\mathrm{\partial <!--\; \partial \; -->}x}\mathrm{d}{X}_t+\frac{1}{2}\frac{{\mathrm{\partial <!--\; \partial \; -->}}^{2}f}{{\mathrm{\partial <!--\; \partial \; -->}}^{2}x}(\mathrm{d}{X}_t{)}^2}$
He explained that the notation $(\mathrm{d}{X}_t{)}^2$ is to be interpreted as the taking the quadratic variation whenever the algebra would suggest you multiply the differentials. For example, $\mathrm{d}{B}_t\mathrm{d}{B}_t=\mathrm{d}⟨<!--\; ⟨\; -->B_t,B_t{⟩<!--\; ⟩\; -->}_T=\mathrm{d}(T)\text{a.s.}=\mathrm{d}t$ (this last step has its own answer on s.e.) Why is the formal multiplication of (stochastic) differentials interpreted as the quadratic variation?

For which numbers a,b,c and d will the function $f(x)=\frac{(ax+b)}{(cx+d)}$ satisfy $f(f(x))=x$ for all x?

I've got a problem with factorising the denominator of an LTI system. The system is a simple boost converter and my current transfer function is
$$U_O=\frac{{\mathrm{\lambda <!--\; \lambda \; -->}}^{\prime }\cdot <!--\; \cdot \; -->U_I-<!--\; -\; -->I_O\cdot <!--\; \cdot \; -->L\cdot <!--\; \cdot \; -->s}{C\cdot <!--\; \cdot \; -->L\cdot <!--\; \cdot \; -->s^2+{\mathrm{\lambda <!--\; \lambda \; -->}}^{\prime 2}}.$$
I want to solve for the systems poles next and and bring them into time constant representation. I thus use the common approach
$$C\cdot <!--\; \cdot \; -->L\cdot <!--\; \cdot \; -->s^2+{\mathrm{\lambda <!--\; \lambda \; -->}}^{\prime 2}\stackrel{!}{=}0$$
which can be solved with the quadratic formula
$$s=\pm <!--\; \pm \; -->\frac{\sqrt{-<!--\; -\; -->4\cdot <!--\; \cdot \; -->C\cdot <!--\; \cdot \; -->L\cdot <!--\; \cdot \; -->{\mathrm{\lambda <!--\; \lambda \; -->}}^{\prime 2}}}{2\cdot <!--\; \cdot \; -->C\cdot <!--\; \cdot \; -->L}=\pm <!--\; \pm \; -->\frac{{\mathrm{\lambda <!--\; \lambda \; -->}}^{\prime }}{\underset{T_{CL}}{\underset{⏟<!--\; ⏟\; -->}{\sqrt{C\cdot <!--\; \cdot \; -->L}}}}j=\frac{{\mathrm{\lambda <!--\; \lambda \; -->}}^{\prime }}{T_{CL}}j$$
and leads to the factorised form
$$(s+\frac{{\mathrm{\lambda <!--\; \lambda \; -->}}^{\prime }}{T_{CL}}j)\cdot <!--\; \cdot \; -->(s-<!--\; -\; -->\frac{{\mathrm{\lambda <!--\; \lambda \; -->}}^{\prime }}{T_{CL}}j).$$
This result is obviously wrong since expanding the term doesn't give the same result and I'm not able to wrap my head around why this is the case.
Factorising the denominator with the third binomial formula gives the correct factors instead
$$(T_{CL}\cdot <!--\; \cdot \; -->s+{\mathrm{\lambda <!--\; \lambda \; -->}}^{\prime }j)\cdot <!--\; \cdot \; -->(T_{CL}\cdot <!--\; \cdot \; -->s-<!--\; -\; -->{\mathrm{\lambda <!--\; \lambda \; -->}}^{\prime }j)$$
and these are already in time constant representation, which leads to the fully factorised transfer function in time constant representation
$$U_O=\frac{{\mathrm{\lambda <!--\; \lambda \; -->}}^{\prime }\cdot <!--\; \cdot \; -->U_I-<!--\; -\; -->I_O\cdot <!--\; \cdot \; -->L\cdot <!--\; \cdot \; -->s}{{\mathrm{\lambda <!--\; \lambda \; -->}}^{\prime 2}\cdot <!--\; \cdot \; -->\underset{p}{\underset{⏟<!--\; ⏟\; -->}{(\frac{T_{CL}}{{\mathrm{\lambda <!--\; \lambda \; -->}}^{\prime }}\cdot <!--\; \cdot \; -->s+j)}}\cdot <!--\; \cdot \; -->\underset{p^{\ast <!--\; \ast \; -->}}{\underset{⏟<!--\; ⏟\; -->}{(\frac{T_{CL}}{{\mathrm{\lambda <!--\; \lambda \; -->}}^{\prime }}\cdot <!--\; \cdot \; -->s-<!--\; -\; -->j)}}}.$$
Where did I mess up the quadratic formula?
My only approach is that the coefficients aren't
$$\begin{array}{rl}a& =C\cdot <!--\; \cdot \; -->L\\ b& =0\\ c& ={\mathrm{\lambda <!--\; \lambda \; -->}}^{\prime 2}\end{array}$$
and I'm not solving for s, but that the coefficients are
$$\begin{array}{rl}a& =1\\ b& =0\\ c& ={\mathrm{\lambda <!--\; \lambda \; -->}}^{\prime 2}\end{array}$$
and I'm solving for $T_{CL}\cdot <!--\; \cdot \; -->s$ with respect to the quadratic equation
$${\underset{x}{\underset{⏟<!--\; ⏟\; -->}{(T_{CL}\cdot <!--\; \cdot \; -->s)}}}^2+{\mathrm{\lambda <!--\; \lambda \; -->}}^{\prime 2}.$$
Please give some advice on how to deal with such situations, I'm seriously confused and couldn't find an answer browsing through the material about quadratic equations. Different problems are far to over represented.

Convert a non-convex QCQP into a convex counterpart
We consider a possibly non-convex QCQP, with nonnegative variable $x\in <!--\; \in \; -->{\mathbb{R}}^n$,
$$\begin{array}{rlrl}& \underset{x}{\text{minimize}}& & f_0(x)\\ & \text{subject to}& & f_i(x)\le <!--\; \le \; -->0,i=1,\dots <!--\; \dots \; -->,m\\ & & & x\ge <!--\; \ge \; -->0\end{array}$$
where $f_i(x)=\frac{1}{2}{x}^T{P}_{i}x+q_i^{T}x+r_i$, with $P_i\in <!--\; \in \; -->{\mathbb{S}}^n$, $q_i\in <!--\; \in \; -->{\mathbb{R}}^n$, and $r_i\in <!--\; \in \; -->\mathbb{R}$, for $i=0,\cdots <!--\; \cdots \; -->,m$. We do not assume that $P_i\ge <!--\; \ge \; -->0$, so this need not to be a convex problem.
Suppose that $q_i\le <!--\; \le \; -->0$ and $P_i$ have non-positive off-diagonal entries. Explain how to reformulate this problem as a convex problem.
What I Have Done
Since both objective function and constraints are possibly non-convex, I consider their common structure, that is $f_i(x)=\frac{1}{2}{x}^T{P}_{i}x+q_i^{T}x+r_i,i=0,1,2,\cdots <!--\; \cdots \; -->,m$. In this structure, $q_i^{T}x+r_i$ part is affine, so it does not cause problem. Then I only consider $x^T{P}_{i}x$.
The first thing that arises in my mind is to decompose $P_i$ into certain form and make $x^T{P}_{i}x$ become something like $y^{T}Qy$ where $Q\ge <!--\; \ge \; -->0$. However, I did not find anything useful (actually it seems that I should not or any non-convex quadratic programming could be solved in this way).
Even though the previous random thought does not seem to work, I still think certain transformation is necessary, and perhaps some non-linear transformation $y_i=\mathrm{\varphi <!--\; \varphi \; -->}(x_i)$. But coming up with this may need some "magic" and up to now have found anything that helps.

Why isn't $y=a(x^2-<!--\; -\; -->Sx+P)$ same with $x^2-<!--\; -\; -->Sx+P$
If we have roots of the function $y=a{x}^2+bx+c$ we can calculate $S=\frac{-<!--\; -\; -->b}{a}$ and also $P=\frac{c}{a}$. Then we know that we can form the function this way:
$$x^2-<!--\; -\; -->Sx+P$$
So on the other side we know that we have the function $f(x)=y$ in different ways:
$$y=a{x}^2+bx+c$$
($\mathrm{\alpha <!--\; \alpha \; -->}$ and $\mathrm{\beta <!--\; \beta \; -->}$ are roots of the quadratic function)
$y=a(x-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->})(x-<!--\; -\; -->\mathrm{\beta <!--\; \beta \; -->})$
And my question is here:
$$y=a(x^2-<!--\; -\; -->Sx+P)$$
Actually know that how we can form the qudratic equation using $x^2-<!--\; -\; -->Sx+P$, but the function must be like $y=a(x^2-<!--\; -\; -->Sx+P)$. Actually I don't know that why we add a. I know it will be removed when $(a)(\frac{-<!--\; -\; -->b}{a})$. But I don't know that what is $y=a(x^2-<!--\; -\; -->Sx+P)$ different whitout a!

Use the quadratic formula to solve the equation.
$$x^2-<!--\; -\; -->4x-<!--\; -\; -->12=0$$

Degree of Equations
A) Which variables in the formula $V=\mathrm{\pi <!--\; \pi \; -->}{r}^{2}h$ would you need to set as a constant in order to generate:
1. a linear equation?
2. a quadratic equation?
B) How should r and h be related to generate a cubic function?

Calculating endpoint for a function so arc length equals l.
I'm trying to simulate a line hanging from a given point using a quadratic function.
The line is located at point $(x_0,y(x_0))$ where y is my quadratic function.
Now, the line has length l, and I simulate it hanging and swinging from side to side by changing the quadratic and linear coefficient. This isn't really important in the problem, but I mention it just to give You a full picture.
Now, the problem is, that when I change the coefficients of the formula, the length of the line changes as well, and I have to pick a point $x_1$ in which the line will end so the length can stay the same.
Of course, the simplest way is just to pick a point $x_1$ such that distance from $(x_0,y(x_0))$ to $(x_1,y(x_1))$ is equal to l, but this actually calculates the straight-line distance, not the arc length, so for certain coefficeints this becomes really inacurate.
So, the proper way to find $x_2$ would be solving this equation:
$${\int <!--\; \int \; -->}_{x_0}^{x_1}\sqrt{1+f^{\prime }(t{)}^2}dt=l$$
For a proper $x_1$. However, after integrating it's really complicated to "extract" $x_1$ from the equation.
Thinking about this, I have found a stupid idea that doesn't work, but I dont know why, and this is the main point of my question.
So, for this integral exists a antiderivative F(x), so this whole equation can be represented as:
$$F(x_1)-<!--\; -\; -->F(x_0)=l$$
So:
$$F(x_1)=l+F(x_0)$$
Now, l is a constant, and $x_0$ is set (I'm just solving for $x_1$), so $F(x_0)$ is a constant as well. This means, that I just can differentiate the whole equation over $x_1$ and get:
$$F^{\prime }(x_1)=0=>\sqrt{1+f^{\prime }(x{)}^2}=0$$
Which is obviously wrong, because it doesn't depend at all on values of l and $x_1$.
So now, I have two questions: 1. Why is my differentiaton step wrong? 2. Is there any way to simplify solving for this integral?

Is there an inverse for a quartic function?
For a programming project, I'm trying to figure out the inverse of a quadratic function: $f(x)=a{x}^4+b{x}^3+c{x}^2+dx+e$. In my case, the numbers a, b, c, d and e are all known constants, but rather annoying non-integers going about five places past the decimal point.
I have a function that returns the result when x is processed through the equation, but what I now need is a function that will return the result back to the original input.
Is there an inverse to the function above, which I would be able to re-use the same constants? If I have to go about finding this solution myself, are there any resources that will explain the steps involved in simple language?
It's been twenty years since my last math class, and I don't recall ever getting into solving fourth degree equations. I've found all sorts of information about trying to solve and find the roots for quartic equations, although I don't quite understand what they're doing, but nothing on a plain old inverse of the function.

Solving Cubic Equations with Lagrange Resolvent?
I'm having difficulties understanding my textbook's decription of solving cubic equations using Lagrange Resolvents and symmetric polynomials.
Here's what I understand:
$$x^3+px-<!--\; -\; -->q=(x-<!--\; -\; -->r)(x-<!--\; -\; -->s)(x-<!--\; -\; -->t)$$
We can also write:
$$\mathrm{\lambda <!--\; \lambda \; -->}=r+ws+w^{2}t$$
$$\mathrm{\mu <!--\; \mu \; -->}=wr+s+w^{2}t$$
where $1,w,w^2$ are the cubic roots of 1. I then understand that ${\mathrm{\lambda <!--\; \lambda \; -->}}^2+{\mathrm{\mu <!--\; \mu \; -->}}^3$ and ${\mathrm{\lambda <!--\; \lambda \; -->}}^3{\mathrm{\mu <!--\; \mu \; -->}}^3$ are symmetric polynomials in r, s, and t. It is also solvable that the elemntary symmetric functions in r, s, t are 0,p,q where
$$r+s+t=0$$
$$rs+rt+st=p$$
$$rst=q$$
The part where I get confused is that the book claims that ${\mathrm{\lambda <!--\; \lambda \; -->}}^3$ and ${\mathrm{\mu <!--\; \mu \; -->}}^3$ are the roots of the quadratic polynomial $q(x)=x^2-<!--\; -\; -->({\mathrm{\lambda <!--\; \lambda \; -->}}^3+{\mathrm{\mu <!--\; \mu \; -->}}^3)x+{\mathrm{\lambda <!--\; \lambda \; -->}}^3{\mathrm{\mu <!--\; \mu \; -->}}^3$, which seems obvious to me. Then they claim you can use the quadratic formula to solve for ${\mathrm{\lambda <!--\; \lambda \; -->}}^3$ and ${\mathrm{\mu <!--\; \mu \; -->}}^3$ in terms of p and q, thus allowing you to solve a system of equations to acquire r,s,t.
How can you use the quadratic formula to "explicitly solve for ${\mathrm{\lambda <!--\; \lambda \; -->}}^3$ and ${\mathrm{\mu <!--\; \mu \; -->}}^3$ in terms of p and q"?

Was the quadratic function derived?
By quadratic function, I mean :
$$f(x)=a{x}^2+bx+c$$
Because the specific constant have a different role, e.g. increasing a has a role of narrowing or inverting the parabola, I'm thinking if this was a derived function/equation. If so, how was this derived?
Additionally, for an even more general equation of a circle for instance:
$$x^2+y^2+Dx+Ey+F=0$$
What would be the role of each of the constants: D, E and F?
And how can I go further into this?

Quadratic Division - How to divide two quadratics?
My studies into graphs and models following examples from Khan Academy has helped me on my goal to learn how to chart and model via the quadratic formula.
However, while I have been successful in understanding addition, subtraction and multiplication of 'g' and 'f':
Say
$f(x)=2x^2+15x-<!--\; -\; -->8$
$g(x)=x^2+10x+16$
I have been unsuccessful in understanding division,
Where f(x)/g(x)
Would one start by grouping the functions into their own families?, for example:
$2x^2/x^2$
$15x/10x$
$-<!--\; -\; -->8/16$
I only came to this conclusion as with multiplication you times as so: $(A+B)\cdot <!--\; \cdot \; -->(C+D)=AC+AD+BC+BD$
However from the example on Khan Ac. this seems wrong, it appears that division departs from any systematic way to find the answer, and that Khans division explanation is not as clear as his others on +, -, and * (appears to me!)
I decided to try and explore division once more in its most simple form to see if I was missing something.
$\frac{2}{3}=\mathrm{0.666666666...}$
$\frac{2}{4}=0.5$
Here clearly, certain division of numbers produces what seems like a simple solution into irrational numbers, If I were to divide a cake the same way I would think of each piece as a ratio, however in decimals these numbers are not clearly defined.
And when one starts dividing multiple factors as with above, how does one find the correct way to divide a quadratic equation, without relying on cheap tricks.

Convex optimization problem to quadratic programming problem
Briefly, have the following problem:
$$\begin{gathered} \underset{i=0}{\overset{n}{\sum <!--\; \sum \; -->}}{a}_i(max[F_i(\stackrel{¯<!--\; ¯\; -->}{x}),0]{)}^2\to <!--\; \to \; -->min, \\ s.t. \\ A\stackrel{¯<!--\; ¯\; -->}{x}\le <!--\; \le \; -->b \end{gathered}$$
where $F(\stackrel{¯<!--\; ¯\; -->}{x})$ is a linear function, $a_i><!--\; >\; -->0$, n is huge comparing to the size of x.
It is possible to write an equal Quadratic Programming problem, such as
$$\begin{gathered} \underset{i=0}{\overset{n}{\sum <!--\; \sum \; -->}}{a}_i(G_i{)}^2\to <!--\; \to \; -->min \\ s.t. \\ {G}_i\ge <!--\; \ge \; -->\mathbf{0},i=\mathrm{0..}n \\ {G}_i\ge <!--\; \ge \; -->F_i(\stackrel{¯<!--\; ¯\; -->}{x})i=\mathrm{0..}n \\ A\stackrel{¯<!--\; ¯\; -->}{x}\le <!--\; \le \; -->b \end{gathered}$$
which can be solved very efficiently with an appropriate numerical method.
Unfortunately in my particular case such conversion doesn't work: it adds a lot of new restrictions, and that appropriate numerical method doesn't converge.
I tried to figure out another equal QPP, which adds fewer new constraints, but nothing came across my mind. Is there another way?

Find f(5), if the graph of the quadratic function $f(x)=a{x}^2+bx+c$ intersects the ordinate axis at point (0;3) and its vertex is at point (2;0)
So I used the vertex form, $y=(x-<!--\; -\; -->2{)}^2+3$, got the quadratic equation and then put 5 instead of x to get the answer, but it's wrong. I think I shouldn't have added 3 in the vertex form but I don't know how else I can solve this

At a time t seconds after it is thrown up in the air, a tomato is at a height of f(t) = -4.9t:2 + 25t + 8 meters.
Give all answers below to two decimal places.
(a) Find the average velocity of the tomato during the first 1.6 seconds?
(b) Find the instantaneous velocity of the tomato at t = 1.6

$$2\cos <!--\; \; -->2x$$ formula.

16 to the power of x-9=1/2

If the discriminant is 1, how many roots does the equation have? And how to solve: $$3x^2-<!--\; -\; -->13x+14$$ through the discriminant?