Question

the image shows a geometric representation of the function $f(x)=x^2 - 2x - 6$ written in standard form. what is this function written in vertex form? $f(x)=(x - 1)^2 - 5$ $f(x)=(x + 1)^2 - 5$ $f(x)=(x + 1)^2 - 7$ $f(x)=(x - 1)^2 - 7$

Explanation:

Step1: Recall vertex form formula

The vertex form of a quadratic function is \( f(x) = a(x - h)^2 + k \), where \((h, k)\) is the vertex. For a quadratic in standard form \( f(x) = x^2 + bx + c \), we can complete the square to convert it to vertex form.

Step2: Complete the square for \( f(x) = x^2 - 2x - 6 \)

First, group the \( x \)-terms: \( f(x) = (x^2 - 2x) - 6 \).
Take half of the coefficient of \( x \) (which is \(-2\)), so \(\frac{-2}{2} = -1\), then square it: \((-1)^2 = 1\).
Add and subtract this value inside the parentheses: \( f(x) = (x^2 - 2x + 1 - 1) - 6 \).
Rewrite the first three terms as a perfect square: \( f(x) = ((x - 1)^2 - 1) - 6 \).
Simplify the constants: \( f(x) = (x - 1)^2 - 1 - 6 = (x - 1)^2 - 7 \)? Wait, no, wait. Wait, original function is \( x^2 - 2x - 6 \). Wait, let's recalculate. Wait, when we complete the square:

\( x^2 - 2x = (x - 1)^2 - 1 \), so substituting back: \( f(x) = (x - 1)^2 - 1 - 6 = (x - 1)^2 - 7 \)? But wait, the options have (x - 1)^2 - 5? Wait, maybe I made a mistake. Wait, the original function is \( f(x) = x^2 - 2x - 6 \)? Wait, no, wait the image says "the function \( f(x) = x^2 - 2x - 6 \) written in standard form"? Wait, no, maybe the original function is \( x^2 - 2x - 5 \)? Wait, no, let's check the options. Wait, the first option is \( (x - 1)^2 - 5 \), the fourth is \( (x - 1)^2 - 7 \). Wait, let's do the completion again.

Wait, let's start over. The standard form is \( f(x) = x^2 - 2x - 6 \)? Wait, no, maybe the original function is \( x^2 - 2x - 5 \)? Wait, no, the image shows the geometric representation. Wait, the tiles: the pink square is \( +x^2 \), two blue rectangles are \( -x \) each, and six blue squares are \( -1 \) each? Wait, no, the blue squares: there are six? Wait, the image has a pink square (\( x^2 \)), two blue rectangles (\( -x \) each), and six blue squares (\( -1 \) each? Wait, no, the blue squares: let's count. The first row: two blue squares (maybe -1 each), second row: two, third row: two. So total six blue squares, each -1? Wait, no, maybe the blue rectangles are -x, and the blue squares are -1. So the area would be \( x^2 - x - x + (1 - 1 - 1 - 1 - 1 - 1) \)? Wait, no, completing the square: \( x^2 - 2x \) can be written as \( (x - 1)^2 - 1 \), so if we have \( x^2 - 2x - 6 \), then it's \( (x - 1)^2 - 1 - 6 = (x - 1)^2 - 7 \), but that's the fourth option. But wait, maybe the original function is \( x^2 - 2x - 5 \)? Wait, no, let's check the options. Wait, the first option is \( (x - 1)^2 - 5 \), which would be \( x^2 - 2x + 1 - 5 = x^2 - 2x - 4 \), no. Wait, no, \( (x - 1)^2 - 5 = x^2 - 2x + 1 - 5 = x^2 - 2x - 4 \). Wait, that's not matching. Wait, maybe the original function is \( x^2 - 2x - 6 \)? Wait, no, maybe I misread the function. Wait, the problem says "the function \( f(x) = x^2 - 2x - 6 \) written in standard form"? Wait, no, maybe it's \( x^2 - 2x - 5 \). Wait, let's check the completion again.

Wait, let's take the function \( f(x) = x^2 - 2x - 6 \). Completing the square:

\( x^2 - 2x = (x - 1)^2 - 1 \), so \( f(x) = (x - 1)^2 - 1 - 6 = (x - 1)^2 - 7 \), which is the fourth option. But the first option is \( (x - 1)^2 - 5 \). Wait, maybe the original function is \( x^2 - 2x - 5 \)? Let's try that. Then \( x^2 - 2x - 5 = (x - 1)^2 - 1 - 5 = (x - 1)^2 - 6 \), no. Wait, maybe the original function is \( x^2 - 2x - 5 \)? No, this is confusing. Wait, maybe the image has a typo, or I misread the function. Wait, the problem says "the function \( f(x) = x^2 - 2x - 6 \) written in standard form"? Wait, no, maybe the function is \( x^2 - 2x - 5 \). Wait, let's check the options again. The first option: \( (x - 1)^2 - 5 \) expands to \( x^2 - 2x + 1 - 5 = x^2 - 2x - 4 \). The second: \( (x + 1)^2 - 5 = x^2 + 2x + 1 - 5 = x^2 + 2x - 4 \). The third: \( (x + 1)^2 - 7 = x^2 + 2x + 1 - 7 = x^2 + 2x - 6 \). The fourth: \( (x - 1)^2 - 7 = x^2 - 2x + 1 - 7 = x^2 - 2x - 6 \). Ah! So the original function is \( f(x) = x^2 - 2x - 6 \), so when we complete the square:

\( x^2 - 2x - 6 = (x^2 - 2x + 1) - 1 - 6 = (x - 1)^2 - 7 \), which is the fourth option: \( f(x) = (x - 1)^2 - 7 \). Wait, but let's check the geometric representation. The pink square is \( x^2 \), two blue rectangles are \( -x \) (so total \( -2x \)), and six blue squares are \( -1 \) (so total \( -6 \))? Wait, no, the blue squares: in the image, there are six? Wait, the first row: two blue squares, second row: two, third row: two. So six blue squares, each -1, so total -6. Then the two blue rectangles are -x each, so total -2x. So the area is \( x^2 - 2x - 6 \), which matches the standard form. Then completing the square: \( x^2 - 2x = (x - 1)^2 - 1 \), so \( f(x) = (x - 1)^2 - 1 - 6 = (x - 1)^2 - 7 \), which is the fourth option. Wait, but the options are:

1. \( f(x) = (x - 1)^2 - 5 \)

2. \( f(x) = (x + 1)^2 - 5 \)

3. \( f(x) = (x + 1)^2 - 7 \)

4. \( f(x) = (x - 1)^2 - 7 \)

So the correct answer is the fourth option: \( f(x) = (x - 1)^2 - 7 \). Wait, but let's verify by expanding:

\( (x - 1)^2 - 7 = x^2 - 2x + 1 - 7 = x^2 - 2x - 6 \), which matches the original function. So that's correct.

Answer:

\( f(x) = (x - 1)^2 - 7 \) (the fourth option, assuming the options are listed as:

A. \( f(x) = (x - 1)^2 - 5 \)

B. \( f(x) = (x + 1)^2 - 5 \)

C. \( f(x) = (x + 1)^2 - 7 \)

D. \( f(x) = (x - 1)^2 - 7 \)

So the answer is D. \( f(x) = (x - 1)^2 - 7 \)