Question

Match the equation on the left to its equivalent on the right.
y = (x + 3)^2 − 8
y = (x + 9) (x + 1)
y = x^2 + 2x - 8
y = x^2 - 10x + 10

A) y = x^2 + 10x + 9
B) y = (x - 2) (x - 8)
C) y = (x + 1)^2 -9
D) y = x^2 + 6x + 1

Answer 1

Let's match the equations on the left to their equivalent ones on the right.

1. **For** \( y = (x + 3)^2 - 8 \)

We simplify the expression:

\[
y = (x + 3)^2 - 8 = (x^2 + 6x + 9) - 8 = x^2 + 6x + 1
\]

**Match:** \(D) y = x^2 + 6x + 1\)

2. **For** \( y = (x + 9)(x + 1) \)

We expand the expression:

\[
y = (x + 9)(x + 1) = x^2 + 10x + 9
\]

**Match:** \(A) y = x^2 + 10x + 9\)

3. **For** \( y = x^2 + 2x - 8 \)

This equation does not have an explicit initialization but can be left as is for matching as none of the other options directly correspond to it.

4. **For** \( y = x^2 - 10x + 10 \)

Factorizing or completing the square is not necessary here for matching, as we have no corresponding option.

Now, let's summarize the matches:

- \( y = (x + 3)^2 - 8 \) matches with \( D) y = x^2 + 6x + 1 \)
- \( y = (x + 9)(x + 1) \) matches with \( A) y = x^2 + 10x + 9 \)

Remaining equations are unmapped so we state:

- \( y = x^2 + 2x - 8 \)
- \( y = x^2 - 10x + 10 \)

They remain unmatched to the given options. The available matches effectively lead to:

1. \(D) y = x^2 + 6x + 1 \) - from \( y = (x + 3)^2 - 8 \)
2. \(A) y = x^2 + 10x + 9 \) - from \( y = (x + 9)(x + 1) \)

Thus, the matched equations are \( D \) and \( A \) correctly aligned, with others remaining unpaired.