If \(\mathrm{y} = (\mathrm{x} – 5)^2 + (\mathrm{x} + 1)^2 – 6\), then y is least when x =

1. Translate the problem requirements: We need to find the value of x that makes \(\mathrm{y = (x – 5)^2 + (x + 1)^2 – 6}\) as small as possible. This is asking for the minimum point of a quadratic function.
2. Expand and simplify the expression: Convert the given expression into standard quadratic form to better understand its structure and identify the minimum.
3. Complete the square or find the vertex: Use the standard form to determine where the quadratic reaches its minimum value by finding the x-coordinate of the vertex.
4. Verify using answer choices: Check our calculated answer against the given options to confirm our solution.

Execution of Strategic Approach

1. Translate the problem requirements

Let's think about what this problem is really asking. We have a mathematical expression \(\mathrm{y = (x – 5)^2 + (x + 1)^2 – 6}\), and we want to find the value of x that makes y as small as possible.

Imagine you're looking at a U-shaped curve (like a smile). This curve has a lowest point somewhere, and we want to find the x-coordinate of that lowest point. This is exactly what "y is least when x = ?" means.

The expression involves two squared terms: \(\mathrm{(x – 5)^2}\) and \(\mathrm{(x + 1)^2}\). Since squares are always non-negative (never negative), both of these terms will be zero or positive. The smallest each squared term can be is zero.

Process Skill: TRANSLATE - Converting the question "y is least when x = ?" into finding the minimum of a quadratic function

2. Expand and simplify the expression

To understand this function better, let's expand the squared terms and combine like terms.

Starting with: \(\mathrm{y = (x – 5)^2 + (x + 1)^2 – 6}\)

First, let's expand \(\mathrm{(x – 5)^2}\):
\(\mathrm{(x – 5)^2 = x^2 – 10x + 25}\)

Next, let's expand \(\mathrm{(x + 1)^2}\):
\(\mathrm{(x + 1)^2 = x^2 + 2x + 1}\)

Now we can substitute these back into our original expression:
\(\mathrm{y = (x^2 – 10x + 25) + (x^2 + 2x + 1) – 6}\)
\(\mathrm{y = x^2 – 10x + 25 + x^2 + 2x + 1 – 6}\)
\(\mathrm{y = 2x^2 – 8x + 20}\)

Now we have y in standard quadratic form: \(\mathrm{y = 2x^2 – 8x + 20}\)

3. Complete the square or find the vertex

For a quadratic in the form \(\mathrm{y = ax^2 + bx + c}\), the minimum (or maximum) occurs at \(\mathrm{x = -b/(2a)}\).

In our case: \(\mathrm{a = 2, b = -8, c = 20}\)

So the x-coordinate of the minimum is:
\(\mathrm{x = -(-8)/(2 × 2) = 8/4 = 2}\)

Let's verify this makes intuitive sense by thinking about our original expression \(\mathrm{y = (x – 5)^2 + (x + 1)^2 – 6}\).

We want to minimize \(\mathrm{(x – 5)^2 + (x + 1)^2}\). This means we want x to be as close as possible to both 5 and -1 simultaneously. The point exactly halfway between 5 and -1 is:
\(\mathrm{(5 + (-1))/2 = 4/2 = 2}\)

This confirms our answer!

Process Skill: SIMPLIFY - Converting to standard form and using the vertex formula

4. Verify using answer choices

Our calculation shows that y is minimized when \(\mathrm{x = 2}\).

Looking at the answer choices:

- A. -2
- B. -1
- C. 0
- D. 2
- E. None of the above

Our answer \(\mathrm{x = 2}\) corresponds exactly to choice D.

Let's double-check by substituting \(\mathrm{x = 2}\) into the original expression:
\(\mathrm{y = (2 – 5)^2 + (2 + 1)^2 – 6}\)
\(\mathrm{y = (-3)^2 + (3)^2 – 6}\)
\(\mathrm{y = 9 + 9 – 6 = 12}\)

We can verify this is indeed the minimum by checking a nearby value, say \(\mathrm{x = 1}\):
\(\mathrm{y = (1 – 5)^2 + (1 + 1)^2 – 6 = (-4)^2 + (2)^2 – 6 = 16 + 4 – 6 = 14}\)

Since 12 < 14, this confirms that \(\mathrm{x = 2}\) gives a smaller value than \(\mathrm{x = 1}\).

Final Answer

The answer is D. 2

When \(\mathrm{x = 2}\), y reaches its minimum value. This occurs because \(\mathrm{x = 2}\) is the point that minimizes the sum of squared distances \(\mathrm{(x – 5)^2 + (x + 1)^2}\), being exactly halfway between the two reference points -1 and 5.

Common Faltering Points

Errors while devising the approach

Faltering Point 1: Misunderstanding what "y is least" means
Students may confuse finding the minimum value of y with finding when y equals zero or some other specific value. They might try to set y = 0 and solve for x, rather than recognizing this is an optimization problem asking for the x-value that minimizes the entire expression.

Faltering Point 2: Attempting to minimize each squared term separately
Students might incorrectly think they need to make both \(\mathrm{(x - 5)^2}\) and \(\mathrm{(x + 1)^2}\) equal to zero simultaneously, leading them to believe there's no solution since x cannot equal both 5 and -1 at the same time. They fail to recognize that we need to minimize the sum of these terms.

Errors while executing the approach

Faltering Point 1: Algebraic expansion errors
When expanding \(\mathrm{(x - 5)^2}\) and \(\mathrm{(x + 1)^2}\), students commonly make sign errors. For example, they might expand \(\mathrm{(x - 5)^2}\) as \(\mathrm{x^2 - 10x - 25}\) instead of \(\mathrm{x^2 - 10x + 25}\), or forget to distribute the negative sign properly in the squared terms.

Faltering Point 2: Incorrect application of the vertex formula
Students often misremember or misapply the vertex formula \(\mathrm{x = -b/(2a)}\). Common errors include using \(\mathrm{x = b/(2a)}\) (forgetting the negative sign) or confusing which coefficients represent 'a' and 'b' in their simplified quadratic expression \(\mathrm{y = 2x^2 - 8x + 20}\).

Faltering Point 3: Arithmetic errors in the midpoint approach
When using the intuitive approach of finding the midpoint between -1 and 5, students might calculate \(\mathrm{(5 + (-1))/2}\) incorrectly, getting \(\mathrm{(5 - 1)/2 = 2}\) by accident (right answer, wrong process) or making errors like getting \(\mathrm{(5 + (-1))/2 = 6/2 = 3}\).

Errors while selecting the answer

No likely faltering points - the calculated value \(\mathrm{x = 2}\) directly matches answer choice D, making selection straightforward.