Quadratic Function Real Life Examples and Applications

Local times and quadratic variation
Suppose we have a continuous local martingale $(M_t{)}_{t\ge <!--\; \ge \; -->0}$, where $M_0=0$. Let $L_t^0$ denote the local time of this process at zero. I would like to prove the simple statement that:
$$inf\{t:L_t^0>0\}=inf\{t:⟨<!--\; ⟨\; -->M{⟩<!--\; ⟩\; -->}_t>0\},$$
where the bracket denotes the quadratic variation process.
My initial idea was to use the occupation density formula, which relates local times and the quadratic variation through the integrals involving non-negative Borel measurable functions. However, the formula integrates against the superscript of L rather than the subscript (that is, one measures the change with respect to the position of the process, not with respect to the time for a fixed position). Therefore, I do not hope that this could be useful.
I am not aware of any other (useful) results connecting the quadratic variation and the local times, so I would appreciate any help. I appreciate the fact that (heuristically) the local time measures the time a process spends at a particular point, with the clock given by the quadratic variation rather than the time indexing the process, but I can't quite get a rigorous proof of this.

Quadratic function question
Find an equation of the quadratic function whose f has zeros -1 and 3 and a maximum value of 8.
I've tried to use intercept form, but I'm not sure wha to do with the maximum value of 8.

Converse linear quadratic optimal control
It is well known that for a linear time invariant system
$$\begin{array}{}\text{(1)}& \stackrel{\dot{}<!--\; \dot{}\; -->}{x}=Ax+Bu\end{array}$$
with (A,B) controllable, there exists a static state feedback $u=-<!--\; -\; -->Kx$ such that the cost function
$$\begin{array}{}\text{(2)}& J={\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}{x}^{T}Qx+u^{T}Rudt\end{array}$$
is minimized, assuming $Q\ge <!--\; \ge \; -->0$ (positive semi-definite) and $R>0$ (positive definite). The gain K is the solution of the algebraic Riccati equation:
$$\begin{array}{rl}0& =A^{T}P+PA-<!--\; -\; -->PB{R}^{-<!--\; -\; -->1}{B}^{T}P+Q\\ K& =R^{-<!--\; -\; -->1}{B}^{T}P\\ P& =P^T\ge <!--\; \ge \; -->0\end{array}$$
known as linear quadratic regulator (LQR). However, I wonder whether the converse also holds?
That is, given a stabilizing $K_s$ (such that $A-<!--\; -\; -->B{K}_s$ is Hurwitz), do there exist matrices $Q\ge <!--\; \ge \; -->0$ and $R>0$ such that $u=-<!--\; -\; -->K_{s}x$ minimizes (2) given (1)? Or put differently:
Question: Is every stabilizing linear state feedback optimal in some sense?

If $f(f(x+y))=f(x^2)+f(y^2)$ then $f(x)=?$ for all integers ($f:\mathbb{Z}\to <!--\; \to \; -->\mathbb{Z}$)
I know how to solve the following problem though: if $f(f(x+y))=f(2x)+2f(y)$ then $f(x)=?$ We can easily analyze that f(x) here (in the second problem) is a linear function. And hence solve it by using the linear equation.
But, As for the main problem (the one I mentioned first) I don't know how to proceed. Should I use the quadratic equation? (my calculation says its a quadratic function)

Prove f(x) is quadratic if $f(2x)=4f(x)$ and f(x) is increasing over positive x
The problem arose in the context of kinetic energy, where it can be proven from symmetry principles that $E(2v)=4E(v)$ without assuming $E=m{v}^2/2$
While one may do further physics from this point to prove the desired result (that E is quadratic in v) -- consider a system with other prime numbers of balls, then do algebra to prove the result for rational scaling in v, then use the fact that there are rational numbers between any two real numbers and assume the function is increasing to prove it for all real scaling -- it seems intuitively obvious from this point immediately, that if E is increasing in v, $E=km{v}^2$.
How would one prove this functional equation?

Conditions for roots of quadratic equation
If I have a quadratic function $f(x)=a{x}^2+bx+c$, what are the conditions that should the numbers a,b and c satisfy so that the equation $f(x)=0$ has real roots $x_1$ and $x_2$ such that $x_1<6<x_2<10$?
My answer is the following system:
$f(6).f(10)<0$ (so that one of the roots is between 6 and 10)
$-<!--\; -\; -->\frac{b}{2a}<10.$
However, there is no such answer in the answer sheet.

Find the values of m for which given equation has real roots
${\sin }^2<!--\; \; -->x-<!--\; -\; -->(m-<!--\; -\; -->3)\sin <!--\; \; -->x+m=0$
So I started by first satisfying $D>=0$ and found that m should range from $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},1]\cup <!--\; \cup \; -->[9,\mathrm{\infty <!--\; \infty \; -->})$
However after this, I found roots by quadratic formula as
$\sin <!--\; \; -->x=(m-<!--\; -\; -->3)/2\pm <!--\; \pm \; -->\sqrt{(m-<!--\; -\; -->1)(m-<!--\; -\; -->9)/4}$
However this value should be within range of sin function and I'm unable to solve the inequality so formed for values of m.
How to solve the inequality( actually the $\pm <!--\; \pm \; -->$ term is confusing me)?

Calculus 2: Partial fractions problem. Finding the value of a constant
I encountered the following problem.
Let f(x) be a quadratic function such that $f(0)=-<!--\; -\; -->6$ and
$$\int <!--\; \int \; -->\frac{f(x)}{x^2(x-<!--\; -\; -->3{)}^8}dx$$
is a rational function.
Determine the value of f′(0)
Here's what I tried. I decomposed the fraction integrand below
$$\frac{f(x)}{x^2(x-<!--\; -\; -->3{)}^8}=\frac{A_1}{x}+\frac{A_2}{x^2}+\underset{i=1}{\overset{8}{\sum <!--\; \sum \; -->}}\frac{B_i}{(x-<!--\; -\; -->3{)}^i}$$
By finding a common denominator, I determined
$$f(x)=A_{1}x(x-<!--\; -\; -->3{)}^8+A_2(x-<!--\; -\; -->3{)}^8+x^2\underset{j=1}{\overset{8}{\sum <!--\; \sum \; -->}}[B_j(x-<!--\; -\; -->3{)}^{8-<!--\; -\; -->j}]$$
$$f^{\prime }(x)=A_1(x-<!--\; -\; -->3{)}^8+8A_2(x-<!--\; -\; -->3{)}^7+D(x)$$
where D(x) is a function such that D(0)=0. (These are all the remaining terms that go away when we plug 0 into f′(0)).I used the information that $D(0)=0$ to get the equation
$$f(0)=A_2(-<!--\; -\; -->3{)}^8=-<!--\; -\; -->6\to <!--\; \to \; -->A_2=\frac{-<!--\; -\; -->2}{3^7}$$
This leaves us with
$$f^{\prime }(0)=A_1(-<!--\; -\; -->3{)}^8+8\cdot <!--\; \cdot \; -->\frac{-<!--\; -\; -->2}{3^7}\cdot <!--\; \cdot \; -->(-<!--\; -\; -->3{)}^7$$
$$f^{\prime }(0)=3^8\cdot <!--\; \cdot \; -->A_1+16$$
I was told $f^{\prime }(0)=16$, however I cannot convince myself the value of $A_1$. I would think that the fact that f(x) is a quadratic function should come into play here. Please let me know what you think.

Transcendental Equation with Quadratic Part (Can it be solved via Lambert W function)?
There I hope to minimize an optimization problem:
$$\underset{x\in <!--\; \in \; -->{\mathbb{R}}_{+}}{min}f(x)=x^2-<!--\; -\; -->y{e}^{-<!--\; -\; -->x^2}+r(x-<!--\; -\; -->d{)}^2,$$
where $y,d\in <!--\; \in \; -->\mathbb{R}$ and $r\in <!--\; \in \; -->{\mathbb{R}}_{+}$.
For this equation, the most natural idea is to find the critical point $x_0$ where $f^{\prime }(x_0)=0$. That is, $(r+1)x+yx{e}^{-<!--\; -\; -->x^2}=r$.
But this form is not similar with Lambert W function since quadratic term is involved. Is there any method to find the analytic/ numerical solution of this equation?

Quadratic Function for braking distance
I have this question about finding a quadratic equation:
Car Velocity v (km/hr) : a series of non zero positive values
Car Brake distance d (meter) : also a series of non zero positive values.
The problem asks why a quadratic equation is suitable to find the relationship between v and d?
The problem also asks if $d=a{v}^2+bv+c$ where a, b and c are constants, find the function.
Here are v values (km/hr) : 20, 30, 40, 50, 60, 70, 80, 90, 100 Here are d values (meters) : 15, 20, 25, 40, 50, 65, 80, 100, 115
My observation so far: I have seen a range of x values including zero value and corresponding y values being converted to quadratic equation using 1st difference and 2nd difference of y values. That solution may not work all the time, especially when I don't have zero value for x.
Secondly, I think the problem lacks key information such as deceleration, response time.

I have two functions and want to find out in which points they intersect
$f(x)=-<!--\; -\; -->3x^3-<!--\; -\; -->3x^2+Ax$
$g(x)=x^3+x^2-<!--\; -\; -->6x$
I know that parameter should be $A=18$, but it was just my guess. What I want to as you for is the process how to find it.
I know that I must start by $f(x)=g(x)$ and then $f(x)-<!--\; -\; -->g(x)=0$ and so on, but I am stuck with quadratic equation with parameter.

How to find quadratic function in vertex form from two points?
A WNBA player takes a three-point shot 22 feet away from the basket, The ball reaches its highest point, 16 feet above the floor, when it is 12 feet (horizontal distance) away from the player. As the ball approaches the basket, it is 13.6 feet above the floor when it is 18 feet away from the player.
a. Write a function in vertex form to model the shot of the basketball. (Let x be the horizontal distance from the player and y be the height of the basketball above the floor).
From today's lesson on quadratic formulas in the vertex form, the vertex form is $y=a(x-<!--\; -\; -->h{)}^2+k$, where a is the growth factor, h is the line of symmetry (x-coordinate of vertex), and k is the y-coordinate of the vertex.
From the question, I can get that the vertex is (12,16), so h is 12 and k is 16. So that means the equation is $y=?(x-<!--\; -\; -->12{)}^2+16$. Another point is (18,13.6). I might be able to infer that another point is (22,0), but it wouldn't help much. a is missing.
There is no more information given. My teacher said that many of us would most likely come back tomorrow begging her for answers, so maybe the answer is not purely mathematical, but requires thinking. How do I solve this (for a)?

Find all functions f such that $f(x-<!--\; -\; -->f(y))=f(f(x))-<!--\; -\; -->f(y)-<!--\; -\; -->1$
So far, I've managed to prove that if f is linear, then either $f(x)=x+1$ or $f(x)=-<!--\; -\; -->1$ must be true. I did this by plugging in $x=0$ to the above equation, which yields
$$f(-<!--\; -\; -->f(y))=f(f(0))-<!--\; -\; -->f(y)-<!--\; -\; -->1$$
and plugging in x instead of y and subtracting, this becomes
$$f(-<!--\; -\; -->f(y))-<!--\; -\; -->f(-<!--\; -\; -->f(x))=f(x)-<!--\; -\; -->f(y)$$
Assuming $f(x)=ax+b$ then gives
$$\begin{array}{rl}f(-<!--\; -\; -->ay-<!--\; -\; -->b)-<!--\; -\; -->f(-<!--\; -\; -->ax-<!--\; -\; -->b)=ax-<!--\; -\; -->ay& \Rightarrow <!--\; \Rightarrow \; -->-<!--\; -\; -->a^{2}y-<!--\; -\; -->ab+a^{2}x+ab=a(x-<!--\; -\; -->y)\\ & \Rightarrow <!--\; \Rightarrow \; -->a^2(x-<!--\; -\; -->y)=a(x-<!--\; -\; -->y).\end{array}$$
Thus $a=1$ or $a=0$. If $a=0$, then the original equation becomes $b=b-<!--\; -\; -->b-<!--\; -\; -->1$, thus $b=-<!--\; -\; -->1$. If $a=1$, the original equation becomes
$$x-<!--\; -\; -->y-<!--\; -\; -->b+b=x+2b-<!--\; -\; -->y-<!--\; -\; -->b-<!--\; -\; -->1⟹<!--\; ⟹\; -->b=1.$$
I briefly tried finding a quadratic function that works but didn't find anything. So my question is: how can I either show that f must be linear or find all other representations?

Prove that sinx is not a rational function using the fact that it is not of the form p(x)/q(x) where p and q are polynomials. Then, by using the above proof, prove that sinx does not satisfy a "quadratic equation" of the form:
$$(\sin <!--\; \; -->x{)}^2{f}_2(x)+(\sin <!--\; \; -->x)f_1(x)+f_0(x)=0,$$
where $f_0,f_1,f_2$ are rational functions.
I know that a rational function cannot be zero at infinitely many points unless it is 0 everywhere, but how does one use this information to formulate p(x)/q(x) argument? If anybody could please help.

Solve:
$$(8x+1)(x^2+x+9)-<!--\; -\; -->(x^3-<!--\; -\; -->8)$$

Revenue Quadratic Functions
I am trying to determine a quadratic function to represent the following description.
A travel agenxy offers a group rate of $2400 per person for a week in London if 16 people sign up for the tour. For each additional person who signs up, the price per person is reduced $100. How many people, in total, must sign up for the tour in order for the travel agency to maximize their revenue? Determine the maximum revenue.
I know to solve the number of people would be to find the axis of symmetry, then for the maximum revenue it would be simply plugging that value into the equation to determine the max revenue.
But how do I set up the quadratic equation??

Sturm-Liouville problem - Partial Differential Equations
I was given the following Sturm-Liouville differential equation problem:
$x^2{y}^{″}+5x{y}^{\prime }+\mathrm{\lambda <!--\; \lambda \; -->}y=0$
$y(1)=y(e^{\mathrm{\pi <!--\; \pi \; -->}})=0$
I displayed the equation in the traditional Sturm-Liouville manner and found that the weight function equals $x^3$, but I am having trouble finding the eigenvalues for this problem.
We were told in class that because the coefficients of the ODE are functions of x and not constants, we should guess a solution for y in the form of $x^m$ and continue from there finding the eigenvalues.
Doing that I got a simple quadratic equation involving m:
$m^2-<!--\; -\; -->4m+\mathrm{\lambda <!--\; \lambda \; -->}=0$
But because it is a quadratic equation where p and q are both non zero I can't seem to find a way to divide the eigenvalues into the common categories which we usually did in class (for $\mathrm{\lambda <!--\; \lambda \; -->}<0,>0,=0$), like in this case for instance:
$m^2+\mathrm{\lambda <!--\; \lambda \; -->}=0$
I believe the solution is very simple but just can't find it, help would be much appreciated!

Quadratic Functions
Consider the strictly convex quadratic function $f(x)=\frac{1}{2}{x}^{t}Px-<!--\; -\; -->q^{t}x+r,$, where $P\in <!--\; \in \; -->{\mathbb{R}}^{n\times <!--\; \times \; -->n}$ is a positive definite matrix, $q\in <!--\; \in \; -->{\mathbb{R}}^n$ and $r\in <!--\; \in \; -->\mathbb{R}.$ Let $\mathcal{H}:=\{H:H\text{ is a }k-<!--\; -\; -->\text{dimensional subspace in }{\mathbb{R}}^n\}.$ Clearly, the restriction of f to any $H\in <!--\; \in \; -->\mathcal{H}$ is again a strictly convex function. For any $H\in <!--\; \in \; -->\mathcal{H},$ we will use $x_H$ to denote the unique optimal point of the following problem
$$\underset{x\in <!--\; \in \; -->H}{\text{min.}}f(x).$$
Now consider the map, $\mathrm{\psi <!--\; \psi \; -->}(H)=x_H.$
Prove / Disprove: The map $\mathrm{\psi <!--\; \psi \; -->}$ is bijective.
Remark: It is assumed that P is invertible and $q\ne <!--\; \ne \; -->\mathbf{0}.$

Solving $2-<!--\; -\; -->2{\sin }^2<!--\; \; -->(x)-<!--\; -\; -->\frac{3}{2}\sin <!--\; \; -->(2x)=0$
I need to solve this equation(preparation for exam).
$2-<!--\; -\; -->2{\sin }^2<!--\; \; -->(x)-<!--\; -\; -->\frac{3}{2}\sin <!--\; \; -->(2x)=0$
I have a idea to convert all sin function to "x", and the solve it using quadratic equation, but I am not sure how to do it, because, here I have two different sin function ${\sin }^2<!--\; \; -->(x)$ and sin(2x).

Find function which contains two points
A certain function contains points (-3, 5) and (5, 2). We are asked to find this function,of course this will be simplest if we consider slope form equation
$$y-<!--\; -\; -->y_1=m(x-<!--\; -\; -->x_1)$$
but could we find for general form of equation? for example quadratic? cubic?