How to determine a quadratic function $f(x) = (x-\text{___})^2 + \text{___}$?

Question

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

1. the parabola lies exactly on the $y$ axis (not a parallel of the $y$ axis) and
2. the parabole does not go through the point $(1|1)$.

Where did I make a mistake?

Answer 1

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 $c=5$.

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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.

Answer 2

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 $f(1) = 5$.

Answer 3

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?