Completing the Square: Solving y = x² - 6x - 5

Let's recall the shift rule which states that for the parabola:

$y=a(x-x_1)^2+y_1$

We get from the parabola:

$y=ax^2$

By shifting it (meaning - shifting all points on it)

  x_1

$y_1$ units (positive) to the right, and units (positive) upward,

(We'll also remember that for negative x1 or y1 - then the shift (in positive units) is in the opposite direction according to this rule),

Therefore, we want to express the parabola, after the shift:

$y=x^2-6x-5$

In the form:

$y=(x-x_1)^2+y_1$

For this we'll use completing the square -

Let's first recall the principles of the "completing the square" method and its general concept:

In this method we use the shortened multiplication formulas for binomial squares in order to give an expression the form of a binomial square,

This method is called "completing the square" because in this method we "complete" a missing part to a certain expression in order to get from it a binomial square form,

Meaning we use the shortened formulas for binomial squares:

$(c\pm d)^2=c^2\pm2cd+d^2$

And we'll bring the expression to a square form by adding and subtracting the missing term,

In the given problem we'll first look at the given parabola after the shift:

$y=x^2-6x-5$

First, we'll try to give the parabola expression a form that resembles the right side of the shortened multiplication formulas, we'll also identify that we're interested in the subtraction form of the shortened multiplication formula, since the non-squared term in the given expression, -6X, has a negative sign,we'll continue, first dealing with the two terms with the highest powers in the expression which are on the left side of the equation,

And we'll try to identify the missing term by comparing to the shortened multiplication formula,

For this - first we'll present these terms in a form similar to the form of the first two terms in the shortened multiplication formula:

$y= \underline{x^2-6x}-5 \textcolor{blue}{\leftrightarrow} \underline{ c^2-2cd+d^2 }\\ \downarrow\\ y=\underline{\textcolor{red}{x}^2\stackrel{\downarrow}{-2 }\cdot \textcolor{red}{x}\cdot \textcolor{green}{3}}-5 \textcolor{blue}{\leftrightarrow} \underline{ \textcolor{red}{c}^2\stackrel{\downarrow}{-2 }\textcolor{red}{c}\textcolor{green}{d}\hspace{2pt}\boxed{+\textcolor{green}{d}^2}} \\$

We can notice that compared to the shortened multiplication formula (right side of the blue press in the previous calculation) we're actually making the analogy:

$\begin{cases} x\textcolor{blue}{\leftrightarrow}c\\ 3\textcolor{blue}{\leftrightarrow}d \end{cases}$

Therefore, we identify that if we want to get from these two terms (underlined in the calculation) a binomial square form,

We'll need to add to these two terms the term$3^2$, but we don't want to change the value of the expression in question, so we'll also subtract this term from the expression,

Meaning- we'll add and subtract the term (or expression) we need to "complete" to a binomial square form,

The following calculation demonstrates the "trick" (two lines under the term we added and subtracted from the expression),

Next - we'll put into binomial square form the appropriate expression (demonstrated using colors) and in the final stage we'll further simplify the expression:

$y= x^2-2\cdot x\cdot 3-5\\ (3x)^2-2\cdot x\cdot 3\underline{\underline{+3^2-3^2}}-5\\ y=\textcolor{red}{x}^2-2\cdot \textcolor{red}{x}\cdot \textcolor{green}{3}+\textcolor{green}{3}^2-9-5\\ \downarrow\\ y= (\textcolor{red}{x}-\textcolor{green}3)^2-9-5\\ \downarrow\\ \boxed{y=(x-3)^2-14}$

Therefore- we got the completing the square form for the expression in the parabola after the shift,

We can now compare to the general form in the shift rule that we mentioned earlier:

$y=(x-x_1)^2+y_1 \\ \updownarrow\\ y=(x-3)^2-14$

We identify therefore using the shift rule we mentioned at the beginning of the solution, that we got this parabola from shifting the parabola:

$y=x^2$

3 units right and 14 units down, therefore the correct answer is answer C