Master Algebra 1 Word Problems with our Comprehensive Practice

Given the first term and the common dimerence er an arithmetic sequence how do you find the 52nd term and the explicit formula: ${\displaystyle a_1=25,d=6}$?

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?

Simplify square root $\sqrt{5}(10-<!--\; -\; -->4\sqrt{2})$

Find the integers if the product of two consecutive integers is 182

What is the Fibonacci sequence?

False or true, with $m$ and $\in$?
If $m$ is a rational no negative number and $n$ an irrational number, then $mn$ is irrational.

Write an equation of a line going through (-3,-5) parallel to y=-4x+1

Find the common ratio of −3,6,−12,24,...

How do you find the first three terms if the common ratio in a geometric sequence is 3/2, and the fifth term is 1

Squares of side 3 inches are cut from each corner of a square sheet of paper. The paper is folded to form an open box with a volume of 675 cu in. How long were the sides of the sheet of paper?

How do you find the simple interest paid after 3 months on a loan of $9700 borrowed at an annual interest rate of 11%?

Find two consecutive integers whose sum is 73

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

Write an nth term rule for r=6 and ${\displaystyle a_3=72}$

Prove that among $2n+1$ irrational numbers we can choose $n+1$ numbers such that the sum of any two chosen numbers is irrational.

Prove that this sum of two irrational numbers is rational number
$$\sqrt[3]{2-<!--\; -\; -->\sqrt{5}}+\sqrt[3]{2+\sqrt{5}}$$

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?