QUESTION IMAGE
Question
your team: your team will be assigned its own parabola to study. investigate your teams parabola and be ready to describe everything you can about it by using its graph, rule, and table. answer the questions below to get your investigation started.
| x | $y = -x^2 + 3x + 4$ | y |
|---|---|---|
| -2 | ||
| -1 | ||
| 0 | ||
| 1 | ||
| 2 | ||
| 3 |
(graph of a coordinate plane with axes drawn)
To complete the table for the function \( y = -x^2 + 3x + 4 \) (assuming the intended function, as the original has a typo), we calculate \( y \) for each \( x \):
Step 1: For \( x = -2 \)
Substitute \( x = -2 \) into \( y = -x^2 + 3x + 4 \):
\( y = -(-2)^2 + 3(-2) + 4 = -4 - 6 + 4 = -6 \)
Step 2: For \( x = -1 \)
Substitute \( x = -1 \):
\( y = -(-1)^2 + 3(-1) + 4 = -1 - 3 + 4 = 0 \)
Step 3: For \( x = 0 \)
Substitute \( x = 0 \):
\( y = -(0)^2 + 3(0) + 4 = 0 + 0 + 4 = 4 \)
Step 4: For \( x = 1 \)
Substitute \( x = 1 \):
\( y = -(1)^2 + 3(1) + 4 = -1 + 3 + 4 = 6 \)
Step 5: For \( x = 2 \)
Substitute \( x = 2 \):
\( y = -(2)^2 + 3(2) + 4 = -4 + 6 + 4 = 6 \)
Step 6: For \( x = 3 \)
Substitute \( x = 3 \):
\( y = -(3)^2 + 3(3) + 4 = -9 + 9 + 4 = 4 \)
The completed table is:
| \( x \) | \( y = -x^2 + 3x + 4 \) | \( y \) |
|---|
\( y = -(-3)^2 + 3(-3) + 4 = -9 - 9 + 4 = -14 \). So the initial \( x=-3 \) was miscalculated. Let's redo all with correct function \( y = -x^2 + 3x + 4 \):
| \( x \) | \( y = -x^2 + 3x + 4 \) | \( y \) |
|---|---|---|
| \( -2 \) | \( -(-2)^2 + 3(-2) + 4 \) | \( -4 - 6 + 4 = -6 \) |
| \( -1 \) | \( -(-1)^2 + 3(-1) + 4 \) | \( -1 - 3 + 4 = 0 \) |
| \( 0 \) | \( -(0)^2 + 3(0) + 4 \) | \( 4 \) |
| \( 1 \) | \( -(1)^2 + 3(1) + 4 \) | \( -1 + 3 + 4 = 6 \) |
| \( 2 \) | \( -(2)^2 + 3(2) + 4 \) | \( -4 + 6 + 4 = 6 \) |
| \( 3 \) | \( -(3)^2 + 3(3) + 4 \) | \( -9 + 9 + 4 = 4 \) |
If the intended function was \( y = x^2 + 2 \) (as handwritten), recalculate:
Step 1: \( x = -2 \)
\( y = (-2)^2 + 2 = 4 + 2 = 6 \)
Step 2: \( x = -1 \)
\( y = (-1)^2 + 2 = 1 + 2 = 3 \)
Step 3: \( x = 0 \)
\( y = 0^2 + 2 = 2 \)
Step 4: \( x = 1 \)
\( y = 1^2 + 2 = 3 \)
Step 5: \( x = 2 \)
\( y = 2^2 + 2 = 6 \)
Step 6: \( x = 3 \)
\( y = 3^2 + 2 = 11 \)
Assuming the handwritten \( y = x^2 + 2 \) is correct, the table becomes:
| \( x \) | \( y = x^2 + 2 \) | \( y \) |
|---|---|---|
| \( -2 \) | \( (-2)^2 + 2 \) | \( 6 \) |
| \( -1 \) | \( (-1)^2 + 2 \) | \( 3 \) |
| \( 0 \) | \( 0^2 + 2 \) | \( 2 \) |
| \( 1 \) | \( 1^2 + 2 \) | \( 3 \) |
| \( 2 \) | \( 2^2 + 2 \) | \( 6 \) |
| \( 3 \) | \( 3^2 + 2 \) | \( 11 \) |
Final Table (for \( y = x^2 + 2 \)):
| \( x \) | \( y = x^2 + 2 \) | \( y \) |
|---|---|---|
| \( -2 \) | \( 4 + 2 \) | \( 6 \) |
| \( -1 \) | \( 1 + 2 \) | \( 3 \) |
| \( 0 \) | \( 0 + 2 \) | \( 2 \) |
| \( 1 \) | \( 1 + 2 \) | \( 3 \) |
| \( 2 \) | \( 4 + 2 \) | \( 6 \) |
| \( 3 \) | \( 9 + 2 \) | \( 11 \) |
(Note: The original table had a typo in the function. The above uses the handwritten \( y = x^2 + 2 \) for consistency.)
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To complete the table for the function \( y = -x^2 + 3x + 4 \) (assuming the intended function, as the original has a typo), we calculate \( y \) for each \( x \):
Step 1: For \( x = -2 \)
Substitute \( x = -2 \) into \( y = -x^2 + 3x + 4 \):
\( y = -(-2)^2 + 3(-2) + 4 = -4 - 6 + 4 = -6 \)
Step 2: For \( x = -1 \)
Substitute \( x = -1 \):
\( y = -(-1)^2 + 3(-1) + 4 = -1 - 3 + 4 = 0 \)
Step 3: For \( x = 0 \)
Substitute \( x = 0 \):
\( y = -(0)^2 + 3(0) + 4 = 0 + 0 + 4 = 4 \)
Step 4: For \( x = 1 \)
Substitute \( x = 1 \):
\( y = -(1)^2 + 3(1) + 4 = -1 + 3 + 4 = 6 \)
Step 5: For \( x = 2 \)
Substitute \( x = 2 \):
\( y = -(2)^2 + 3(2) + 4 = -4 + 6 + 4 = 6 \)
Step 6: For \( x = 3 \)
Substitute \( x = 3 \):
\( y = -(3)^2 + 3(3) + 4 = -9 + 9 + 4 = 4 \)
The completed table is:
| \( x \) | \( y = -x^2 + 3x + 4 \) | \( y \) |
|---|
\( y = -(-3)^2 + 3(-3) + 4 = -9 - 9 + 4 = -14 \). So the initial \( x=-3 \) was miscalculated. Let's redo all with correct function \( y = -x^2 + 3x + 4 \):
| \( x \) | \( y = -x^2 + 3x + 4 \) | \( y \) |
|---|---|---|
| \( -2 \) | \( -(-2)^2 + 3(-2) + 4 \) | \( -4 - 6 + 4 = -6 \) |
| \( -1 \) | \( -(-1)^2 + 3(-1) + 4 \) | \( -1 - 3 + 4 = 0 \) |
| \( 0 \) | \( -(0)^2 + 3(0) + 4 \) | \( 4 \) |
| \( 1 \) | \( -(1)^2 + 3(1) + 4 \) | \( -1 + 3 + 4 = 6 \) |
| \( 2 \) | \( -(2)^2 + 3(2) + 4 \) | \( -4 + 6 + 4 = 6 \) |
| \( 3 \) | \( -(3)^2 + 3(3) + 4 \) | \( -9 + 9 + 4 = 4 \) |
If the intended function was \( y = x^2 + 2 \) (as handwritten), recalculate:
Step 1: \( x = -2 \)
\( y = (-2)^2 + 2 = 4 + 2 = 6 \)
Step 2: \( x = -1 \)
\( y = (-1)^2 + 2 = 1 + 2 = 3 \)
Step 3: \( x = 0 \)
\( y = 0^2 + 2 = 2 \)
Step 4: \( x = 1 \)
\( y = 1^2 + 2 = 3 \)
Step 5: \( x = 2 \)
\( y = 2^2 + 2 = 6 \)
Step 6: \( x = 3 \)
\( y = 3^2 + 2 = 11 \)
Assuming the handwritten \( y = x^2 + 2 \) is correct, the table becomes:
| \( x \) | \( y = x^2 + 2 \) | \( y \) |
|---|---|---|
| \( -2 \) | \( (-2)^2 + 2 \) | \( 6 \) |
| \( -1 \) | \( (-1)^2 + 2 \) | \( 3 \) |
| \( 0 \) | \( 0^2 + 2 \) | \( 2 \) |
| \( 1 \) | \( 1^2 + 2 \) | \( 3 \) |
| \( 2 \) | \( 2^2 + 2 \) | \( 6 \) |
| \( 3 \) | \( 3^2 + 2 \) | \( 11 \) |
Final Table (for \( y = x^2 + 2 \)):
| \( x \) | \( y = x^2 + 2 \) | \( y \) |
|---|---|---|
| \( -2 \) | \( 4 + 2 \) | \( 6 \) |
| \( -1 \) | \( 1 + 2 \) | \( 3 \) |
| \( 0 \) | \( 0 + 2 \) | \( 2 \) |
| \( 1 \) | \( 1 + 2 \) | \( 3 \) |
| \( 2 \) | \( 4 + 2 \) | \( 6 \) |
| \( 3 \) | \( 9 + 2 \) | \( 11 \) |
(Note: The original table had a typo in the function. The above uses the handwritten \( y = x^2 + 2 \) for consistency.)