How to determine a quadratic function $f(x) = (x-\text{___})^2 + \text{___}$? I am helping my child with his homework. I don't know how to solve the following problem:

Determine the equation of the quadratic function $f$ by filling in blanks below.
$f(x) = (x-\text{___})^2 + \text{___}$
The graph of $f$ is symmetrical to the parallel of the y-axis through (1|1). The
  y coordinate of the vertex is 5.

I assume the parabola looks like the red or blue one in the drawing below (please correct me, if I'm wrong).

I assume that the text through (1|1) means that the parabola goes through this point, i. e. $f(1)=1$ (this is also not 100% clear).
According to his theoretical materials, 


*

*the vertex form looks like $f(x) = a\cdot(x-d)^2 + e$ such that $a, d, e \in \mathbb{R}$ and $a \neq 0$ and

*the coordinates of the vertex are $S(-d|e)$.


Because the $y$ coordinate of the vertex is 5, the vertex is $S(-d|5)$. $a=1$ because we are only allowed to fill in the blanks.
$f(-d)=(x-d)^2+e$
$f(-d)=(x-d)^2+5$
I replace $x$ by $-d$.
$f(-d)=(-d-d)^2+5$
I replace $f(-d)$ by $5$ ($y$ coordinate of the vertex).
$5=(-d-d)^2+5$
$5=(-2d)^2+5$
$5=4d^2+5$
$0=4d^2$
$0=d^2$
$d=0$
Then, the equation is $f(x)=(x-0)^2 + 5$.
The graph looks like this:

It is wrong because


*

*the parabola lies exactly on the $y$ axis (not a parallel of the $y$ axis) and

*the parabole does not go through the point $(1|1)$.


Where did I make a mistake?
 A: 
The graph of $f$ is symmetrical to the parallel of the $y$-axis through $(1|1)$.

Although the phrase "symmetrical to" is rather unusual, I cannot imagine anything being meant other than

The graph of $f$ is symmetrical with respect to the line through $(1,1)$ parallel to the $y$-axis.

This implies that $f=(x-1)^2+c$ for some constant $c$. Given that $f(1)=5$ it follows that $c=5$.

As for where you went wrong; you interpret the question as saying that the point $(1,1)$ is on the parabola. But the question clearly states that the vertex is at $(1,5)$, meaning that $f(1)=5$. This makes $f(1)=1$ impossible.
A: For a general function $f(x)$ and $a,b > 0$, $f(x-a) + b$ represents the original graph shifted $a$ units to the right and $b$ units upwards.
Since the equation with blanks has a positive $x$ coefficient (1), your red graph is the one of interest.

I assume that the text through (1|1) means that the parabola goes through this 
  point, i. e. (1)=1 (this is also not 100% clear).

This is not quite correct. The vertex is at $(1, 5)$ meaning $f(1) = 5$.
A: The form in question:
$$f(x) = (x - \alpha)^2 + \beta$$
can be understood as the result of applying two simple geometric transformations to the basic parabola function
$$g(x) := x^2$$
which has symmetry axis at the $y$-axis exactly, and vertex at $(0, 0)$. One of these transformations is a transformation in the $x$-coordinate, which shifts it left or right. This is what $\alpha$ controls: a positive $\alpha$ means shifting to the right, negative, to the left, by that many units of distance. The other transformation is in the $y$-coordinate, which shifts it up or down. That is what $\beta$ controls: positive is up, negative is down, again, by so many units of distance.
Following that, you should wonder: how much do I need to shift the basic parabola left/right so its axis of symmetry lies on $(1, 1)$, and then after doing that, how much up/down so that its vertex comes to lie at $y = 5$?
So for your problem: To get the axis of symmetry to where it should be, do you need a left/right shift of the basic parabola? Does it make sense the amount of this shifting is $\alpha = 0$ units, that is, no shift at all, which is what you have given?
