Quadratic Function Real Life Examples and Applications

Find a solution for a differential equation, assume that $y(x)=a{x}^2+bx+c$
Find a solution to the given equation:
$y^{\prime }+2y=x^2+4x+7$
and we are told to assume that y(x) is a quadratic function, which follows this general form
$y=a{x}^2+bx+c.$
Now to find the solutions, at first I tried to solve the given for y and then see what I could do from there but, I realized that you can't take the derivative of a function that contains its own derivative.
So then I thought that I had to create my own function that follows the perviously mentioned parameters (the given and the quad. form). This is were I am stuck, I can not mathematically deduce a solution for this problem.

Find the form of f(x, y) knowing the form of contour lines in the XZ and YZ planes
I am trying to find the form of $z=f(x,y)$. I know that:
Contour lines in the XZ plane are of the form:
$z=A\ast <!--\; \ast \; -->ln(x)+B$
(the A and B parameters vary with y)
Contour lines in the YZ plane are of the form:
$z=C{y}^2+Dy+E$
(the C, D and E parameters vary with x)
What is then the form of $z=f(x,y)$?

Location of Roots in Quadratic Equation
We learnt about the location of roots of quadratic equation in class today. I have a problem in one case specifically, the one where both the roots are required in the interval $(k_1,k_2)$ where $k_1,k_2$ are real constant numbers. So the conditions given to us were,
1. $\mathrm{\Delta <!--\; \Delta \; -->}$ greater than or equal to 0 (fairly obvious)
2. $af(k_1)>0$ where $f(k_1)$ is the value of the function at $k_1$ and a is the leading coefficient
3. $af(k_2)>0$ (both of these are again understandable)
4. then comes $k_1<-<!--\; -\; -->b/2a<k_2$ i.e. the abscissa of the vertex of the parabola lies in the interval required.

How do I get the tangent coordinates if I am given the quadratic function and the slope(gradient)
Calculate the coordinates of the intersections point between a straight line with a given slope and a quadratic function, so that you only receive one intersection instead of the normal two or none. I am given the slope gradient m and the quadratic equation.
In this example its
$y=x^2+3x-<!--\; -\; -->2,m=1$

Lyapunov equation for stability analysis - what's the point?
In the following theorem $A,P,Q\in <!--\; \in \; -->{\mathbb{R}}^{n\times <!--\; \times \; -->n}$, and P and Q are symmetric. The notation $P>0$ means that the matrix P is positive definite.
Given any $Q>0$, there exists a unique $P>0$ satisfying $A^{T}P+PA+Q=0$ if and only if the linear system $\stackrel{\dot{}<!--\; \dot{}\; -->}{x}=Ax$ is globally asymptotically stable. The quadratic function $V(z)=z^{T}Pz$ is a Lyapunov function that can be used to verify stability.
Recently I became quite familiar with tools to solve Lyapunov equation $A^{T}P+PA+Q=0$ and obtain P. However in the context of stability analysis I see one huge problem. It's that you still have to check that P is positive definite or equivalently that its eigenvalues are positive. But for the linear system the fact that all eigenvalues of A have negative real parts ensures that the system is globally asymptotically stable in the first place.
So what's the point of going through all the hassle just to end up with looking for the eigenvalues (P and A are of the same size!) anyway? Does the symmetric matrix make such a big difference?

Is there a way to analytically find a stationary point along an arbitrary line in a multivariable quadratic function?
Let's say I'm working with a quadratic function with an equation of $f(\mathbf{x})=\frac{1}{2}{\mathbf{x}}^{T}A\mathbf{x}-<!--\; -\; -->b^T\mathbf{x}$. Now, let's take a direction p and transform the function into one a one dimensional one such that $f^{\prime }(\mathrm{\alpha <!--\; \alpha \; -->})=f({\mathbf{x}}_0+\mathrm{\alpha <!--\; \alpha \; -->}\mathbf{p}\mathbf{)}$, where ${\mathbf{x}}_0$ is some starting point.
Is there a way to analytically find an $\mathrm{\alpha <!--\; \alpha \; -->}$ which minimizes or maximizes $f^{\prime }(\mathrm{\alpha <!--\; \alpha \; -->})$? What is instead of arbitrary direction p we have $\mathbf{p}=-<!--\; -\; -->\mathrm{\nabla <!--\; \nabla \; -->}f({\mathbf{x}}_0)$?
(Yes, this is for an optimization problem)

Finding the roots of $(x^2+7x+6{)}^2+7(x^2+7x+6)+6=x$
I have been trying to solve this one problem from the Duke Math Meet, which does not provide a solution:
Find all solutions of $(x^2+7x+6{)}^2+7(x^2+7x+6)+6=x$
At first I tried to factorize the polynomial, but always had the right-hand side x remain, which was inconvenient. I then tried using the fact that one can write the left-hand side as a composition of functions, and equated that with the inverse of the quadratic plugged into the function, but that was a very nasty equation with square roots, still ending up with a quartic.
What other solution paths are viable for this problem, and is there a way to factor the quartic?

Now building a new tunnel that has a shape of parabolic curve. The tunnel is 10 m wide and at 4 m from either side, the height of the tunnel is 6 m. Find the quadratic equation in standard form that models the ceiling of the new tunnel.
The question is very simple, but I just can't understand what does at 4 m from either side mean in this question while mentioning the width in 10m?
I am not a native English speaker so it is hard for me understand what the information this question is asking.
Any comments and answers will be appreicated!

Diophantine Inequality $4ax-<!--\; -\; -->y^2+z<0$
For fixed positive integer a and positive z in
$4ax-<!--\; -\; -->y^2+z<0,$
find how many integer solutions (x,y) there are as a function of $z\in <!--\; \in \; -->{\mathbb{Z}}_{+}$, subject to constraints $x\ge <!--\; \ge \; -->0$ and $a\ge <!--\; \ge \; -->y\ge <!--\; \ge \; -->0$.
This form comes from the quadratic polynomial discriminant. I am not sure where to begin as this isn't the usual quadratic Diophantine inequality (or equation for that matter) that I find through search engines.

Find the range of values of k such that f(x) is onto
I am trying to solve the below problem in pre-calculus, I give below the steps I followed , I am stuck as it gets very complicated , need help in solution.
A function $f:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ is defined by
$f(x)=\frac{k{x}^2+6x-<!--\; -\; -->8}{k+6x-<!--\; -\; -->8x^2}.$
Find the intervals of values of k such that f is onto. The answer given in the book is $2\le <!--\; \le \; -->k\le <!--\; \le \; -->14$.
I started by trying to find the range of the function and find the values of k such that the range is = R ( that is range = co domain , in this case co domain is R and hence has to prove that range = R)
Let
$y=\frac{k{x}^2+6x-<!--\; -\; -->8}{k+6x-<!--\; -\; -->8x^2}$
Rearranging and grouping gives
$(k+8y)x^2+6(1-<!--\; -\; -->y)x-<!--\; -\; -->(8+ky)=0.$
This is quadratic in x and for x to be real $det(x)\ge <!--\; \ge \; -->0$, which requires
$36(1-<!--\; -\; -->y{)}^2+4(k+8y)(8+ky)\ge <!--\; \ge \; -->0.$
Rearranging and grouping this equation gives
$(36+32k)y^2+(4k^2+184)y+(36+32k)\ge <!--\; \ge \; -->0.$
Now I have to find the range of values of k such that the above expression is $\ge <!--\; \ge \; -->0$. Totally stuck here. Please help. Also, is there any other easier way of arriving at the range of k ?

Vertex Equation of an inverse quadratic function.
I'm working on a graphing web tool using JSXGraph, The user should be able to draw different functions. I was able to allow the user to draw quadratic functions by creating the vertex of the function where the user clicks, and then creating another point up-right of the the vertex. Then I construct the function by obtaining a through the vertex equation
$y=a(x-<!--\; -\; -->h{)}^2+k$
I isolate a, which gives
$a=\frac{y-<!--\; -\; -->k}{x^2-<!--\; -\; -->2xh+h^2}$
As I have the vertex and another point, I obtain a, and then isolating y
$f(x)=a{x}^2-<!--\; -\; -->2ahx+a{h}^2+k$
with that, I obtain b and then c.
The problem is I also need to allow the user to draw inverted parabolas, however I can't find a way to obtain the function given the vertex and a point. I believe i would be able to, with the vertex equation of the inverse quadratic, following a similar procedure as the one above, but I'm not even sure if that exists given the dual nature of the inverse quadratic.
So is there a vertex equation for the inverse of a quadratic function that would allow me to accomplish what I did above?

Quadratic Manipulation
Is there some easy way to transform a quadratic equation like $a{x}^2+bx+c$ into a quadratic equation of the form $d(x+s{)}^2+e(x+s)$ where a,b,c,d,e are constants and $a,d,e>0$
-s is a root of the function

Generalizing $x^{⊺<!--\; ⊺\; -->}Ax=0$ theorems to $x^{⊺<!--\; ⊺\; -->}A(x)x=0$ forms?
Let $x\in <!--\; \in \; -->{\mathbb{R}}^n$ be any real vector and let $A=A^{⊺<!--\; ⊺\; -->}\in <!--\; \in \; -->{\mathbb{R}}^{n\times <!--\; \times \; -->n}$ be a real symmetric matrix. Then,
$x^{⊺<!--\; ⊺\; -->}Ax=0\mathrm{\forall <!--\; \forall \; -->}x⟹<!--\; ⟹\; -->A\equiv <!--\; \equiv \; -->0$
Can we still say this if $A=A^{⊺<!--\; ⊺\; -->}:{\mathbb{R}}^n\to <!--\; \to \; -->{\mathbb{R}}^{n\times <!--\; \times \; -->n}$ is a real symmetric matrix-valued function of x? I.e.,
$x^{⊺<!--\; ⊺\; -->}A(x)x=0\mathrm{\forall <!--\; \forall \; -->}x\stackrel{?}{⟹<!--\; ⟹\; -->}A\equiv <!--\; \equiv \; -->0\mathrm{\forall <!--\; \forall \; -->}x$
Also, is there a name for equations of the form $f(x)=x^{⊺<!--\; ⊺\; -->}A(x)x$ with symmetric A? It is obviously not quadratic since A depends on x, but it still has a certain quadratic-form feel.

Limit of sequence to infinity
For a sequence $\{a_n\}$, given that the value of
$\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}(a{}_n{}^2+4n{a}_n+4n^2)=4$
what is the value of
$\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{a_n}{n}$

Equality of a quadratic function
Let $f:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ an arbitrary function and $g:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ a quadratic function with the following property:
For any m and n the equation $f(x)=mx+n$ has a solution iff the equation $g(x)=mx+n$ has a solution.
Prove that f and g are equal.

Converting $a{x}^3+b{x}^2+cx+d$ to $a(x-<!--\; -\; -->j{)}^3+k$
We're all familiar with the vertex form of a quadratic function,
$$a(x-<!--\; -\; -->p{)}^2+q$$
where (-p,q) represent the coordinates of the maximum or minimum point of the parabola. This is achieved by performing completing the square on the standard form of the quadratic function $a{x}^2+bx+c$.
My question is: can the same be done for cubic functions? Playing around with graphs have shown me that
$$a(x-<!--\; -\; -->j{)}^3+k$$
graphs a cubic function whose inflection point is (−j,k). However, I have not been able to find a method of converting a standard cubic function $a{x}^3+b{x}^2+cx+d$ to its 'vertex form' as stated above that is applicable generally to all cubic functions.
I have tried geometrically 'completing the cube' with a friend and found a way (?) to convert standard cubic expressions to their vertex form, but this method was only applicable to cubics without a linear term. I'm looking for a general method that works for all cubics, I really appreciate the help!
Edit: Okay, so I received a lot of feedback which explained why $a(x-<!--\; -\; -->j{)}^3+k$ would not work for all cubic equations because it only has one real root, and got pointers in the direction of depressed cubic equations. My question is: is there a general way to show the position of the inflection point using the depressed cubic?

Is value of $\mathrm{\alpha <!--\; \alpha \; -->}$ defined?
Consider three quadratic functions:
$$P_1(x)=a{x}^2-<!--\; -\; -->bx-<!--\; -\; -->c$$
$$P_2(x)=b{x}^2-<!--\; -\; -->cx-<!--\; -\; -->a$$
$$P_3(x)=c{x}^2-<!--\; -\; -->ax-<!--\; -\; -->b$$
Where $a,b,c\in <!--\; \in \; -->\mathbb{R}\mathrm{\setminus <!--\; \setminus \; -->}\left\{0\right\}$
If there exists real number $\mathrm{\alpha <!--\; \alpha \; -->}$ such that
$$P_1(\mathrm{\alpha <!--\; \alpha \; -->})=P_2(\mathrm{\alpha <!--\; \alpha \; -->})=P_3(\mathrm{\alpha <!--\; \alpha \; -->})$$
Prove that $a=b=c$
My Try:
Using
$$P_1(\mathrm{\alpha <!--\; \alpha \; -->})=P_2(\mathrm{\alpha <!--\; \alpha \; -->})=P_3(\mathrm{\alpha <!--\; \alpha \; -->})$$
we get
$$\begin{array}{}\text{(1)}& (a-<!--\; -\; -->b){\mathrm{\alpha <!--\; \alpha \; -->}}^2-<!--\; -\; -->(b-<!--\; -\; -->c)\mathrm{\alpha <!--\; \alpha \; -->}-<!--\; -\; -->(c-<!--\; -\; -->a)=0\end{array}$$
$$\begin{array}{}\text{(2)}& (b-<!--\; -\; -->c){\mathrm{\alpha <!--\; \alpha \; -->}}^2-<!--\; -\; -->(c-<!--\; -\; -->a)\mathrm{\alpha <!--\; \alpha \; -->}-<!--\; -\; -->(a-<!--\; -\; -->b)=0\end{array}$$
$$\begin{array}{}\text{(3)}& (c-<!--\; -\; -->a){\mathrm{\alpha <!--\; \alpha \; -->}}^2-<!--\; -\; -->(a-<!--\; -\; -->b)\mathrm{\alpha <!--\; \alpha \; -->}-<!--\; -\; -->(b-<!--\; -\; -->c)=0\end{array}$$
Letting $p=a-<!--\; -\; -->b$, $q=b-<!--\; -\; -->c$ and $r=c-<!--\; -\; -->a$ we get the above three equations as:
$$p{\mathrm{\alpha <!--\; \alpha \; -->}}^2-<!--\; -\; -->q\mathrm{\alpha <!--\; \alpha \; -->}-<!--\; -\; -->r=0$$
$$q{\mathrm{\alpha <!--\; \alpha \; -->}}^2-<!--\; -\; -->r\mathrm{\alpha <!--\; \alpha \; -->}-<!--\; -\; -->p=0$$
$$r{\mathrm{\alpha <!--\; \alpha \; -->}}^2-<!--\; -\; -->p\mathrm{\alpha <!--\; \alpha \; -->}-<!--\; -\; -->q=0$$
Using $p+q+r=0$ and Eliminating $\mathrm{\alpha <!--\; \alpha \; -->}$ from last two equations above we get:
$$\begin{array}{}\text{(4)}& \mathrm{\alpha <!--\; \alpha \; -->}=\frac{p^2+pr+r^2}{p^2+pr+r^2}=1\end{array}$$
Substituting $\mathrm{\alpha <!--\; \alpha \; -->}=1$ we get:
$$p-<!--\; -\; -->q-<!--\; -\; -->r=0$$
$⟹<!--\; ⟹\; -->$
$$p=0$$
Similarly
$$q=r=0$$
But $p=r=0$ is contradicting the result in (4)
So is $\mathrm{\alpha <!--\; \alpha \; -->}=1$ or $\mathrm{\alpha <!--\; \alpha \; -->}$ is not defined?

A free website which can solve this equation
solve for v, $h(v-<!--\; -\; -->t)=h(v+t)$ where $h(x)=a{x}^2+bx+c$. The value is the vertex of the quadratic function the website wolfarmalpha can solve that when a,b and c are numbers. I want the students to figure the vertex of the quadratic function.

Finding the graph of a function based on its properties
I know the following things about my function: $f(–4)=f(2)=0;f^{\prime }(-<!--\; -\; -->1)=0,f^{\prime }(x)<0\text{ for }x<-<!--\; -\; -->1;f^{\prime }(x)>0\text{ for }x>-<!--\; -\; -->1$
That is, it appears to be a quadratic function with intercepts at 2 and -4.
How can I determine its equation?

How to figure out the formula of axis of simmetry in quadratic functions
Well, this is what I did. Assuming that that quadratic equation or function is given by the general formula $f(x)=a{x}^2+bx+c$.
So, assuming that there exist an axis of simmetry (eje de simetria en español), I had to find two points $(x_1,f(x_1))$ and $(x_2,f(x_2))$ so that $x_1\ne <!--\; \ne \; -->x_2$ but $f(x_1)=f(x_2)$. To make the typing job a bit easier $x_1=x$ and $x_2=d$.
So $f(x)=a{x}^2+bx+c$ and $f(d)=a{d}^2+bd+c$ and $f(x)=f(d)$:
$a{x}^2+bx+c=a{d}^2+bd+c$
$a{x}^2+bx=a{d}^2+bd$
$a{x}^2-<!--\; -\; -->a{d}^2=bd-<!--\; -\; -->bx$
$a(x^2-<!--\; -\; -->d^2)=b(d-<!--\; -\; -->x)$
$a(x+d)(x-<!--\; -\; -->d)=-<!--\; -\; -->(x-<!--\; -\; -->d)b$
$a(x+d)=-<!--\; -\; -->b$
Because we are talking about quadratic functions b can be 0. So if we $b=0$, we have:
$a(x+d)=0$
and again, if it is a quadractic function a can´t be 0, so we´d have that:
$x+d=0$ and $x=-<!--\; -\; -->d$, which is true. For example in the function $f(x)=x^2-<!--\; -\; -->16$, $f(4)=f(-<!--\; -\; -->4)=0$.
Well, continuing with the calculations we´d have:
$x+d=\frac{-<!--\; -\; -->b}{a}$
$d=\frac{-<!--\; -\; -->b}{a}-<!--\; -\; -->x$
$d=\frac{-<!--\; -\; -->b-<!--\; -\; -->xa}{a}$
Because it´s an axis of simmetry, it´s in the middle, so I´ll have to use this formula:
$\frac{x_1+x_2}{2}$
So: $x_1+x_2=x+d$ and $d=\frac{-<!--\; -\; -->b-<!--\; -\; -->xa}{a}$, so I would have.
$x+d=x+(\frac{-<!--\; -\; -->b-<!--\; -\; -->xa}{a})$
$=\frac{xa}{a}+(\frac{-<!--\; -\; -->b-<!--\; -\; -->xa}{a})$
$=\frac{xa-<!--\; -\; -->b-<!--\; -\; -->xa}{a}$
$x+d=\frac{-<!--\; -\; -->b}{a}$
So:
$\frac{x+d}{2}=\frac{-<!--\; -\; -->b}{a}÷2$
$=\frac{-<!--\; -\; -->b}{2a}$
That´s what I´ve done so far, but what I probably want to know is why there exists and axis of symmetry (which I think will be the line $x=\frac{-<!--\; -\; -->b}{2a}$). I thought I could prove it using that fact but I got nowhere that´s why I posted this forum. (there might be some grammar mistakes, that´s because I am not an english speaker, you can correct where you think I made a mistake.