Question

the table shows the corresponding values of x and y: when x is -2, y is -5; when x is -1, y is -2; when x is 0, y is 3; when x is 1, y is 10; when x is 2, y is 19.

Explanation:

To determine the relationship between \( x \) and \( y \), we can assume a quadratic function of the form \( y = ax^2 + bx + c \) (since the differences of differences might be constant, indicating a quadratic relationship).

Step 1: Use the point \( (0, 3) \)

When \( x = 0 \), \( y = 3 \). Substituting into \( y = ax^2 + bx + c \):
\[
3 = a(0)^2 + b(0) + c \implies c = 3
\]

Step 2: Use the point \( (-1, -2) \)

Substitute \( x = -1 \), \( y = -2 \), and \( c = 3 \) into \( y = ax^2 + bx + c \):
\[
-2 = a(-1)^2 + b(-1) + 3 \implies -2 = a - b + 3 \implies a - b = -5 \quad (1)
\]

Step 3: Use the point \( (1, 10) \)

Substitute \( x = 1 \), \( y = 10 \), and \( c = 3 \) into \( y = ax^2 + bx + c \):
\[
10 = a(1)^2 + b(1) + 3 \implies 10 = a + b + 3 \implies a + b = 7 \quad (2)
\]

Step 4: Solve the system of equations

We have the system:
\[

$$\begin{cases} a - b = -5 \\ a + b = 7 \end{cases}$$

\]
Add the two equations:
\[
(a - b) + (a + b) = -5 + 7 \implies 2a = 2 \implies a = 1
\]
Substitute \( a = 1 \) into equation (2):
\[
1 + b = 7 \implies b = 6
\]

Step 5: Verify the function

The quadratic function is \( y = x^2 + 6x + 3 \). Let's verify with \( x = -2 \):
\[
y = (-2)^2 + 6(-2) + 3 = 4 - 12 + 3 = -5
\]
Which matches the table. For \( x = 2 \):
\[
y = (2)^2 + 6(2) + 3 = 4 + 12 + 3 = 19
\]
Which also matches the table.

So the relationship between \( x \) and \( y \) is \( \boldsymbol{y = x^2 + 6x + 3} \).

Answer:

To determine the relationship between \( x \) and \( y \), we can assume a quadratic function of the form \( y = ax^2 + bx + c \) (since the differences of differences might be constant, indicating a quadratic relationship).

Step 1: Use the point \( (0, 3) \)

When \( x = 0 \), \( y = 3 \). Substituting into \( y = ax^2 + bx + c \):
\[
3 = a(0)^2 + b(0) + c \implies c = 3
\]

Step 2: Use the point \( (-1, -2) \)

Substitute \( x = -1 \), \( y = -2 \), and \( c = 3 \) into \( y = ax^2 + bx + c \):
\[
-2 = a(-1)^2 + b(-1) + 3 \implies -2 = a - b + 3 \implies a - b = -5 \quad (1)
\]

Step 3: Use the point \( (1, 10) \)

Substitute \( x = 1 \), \( y = 10 \), and \( c = 3 \) into \( y = ax^2 + bx + c \):
\[
10 = a(1)^2 + b(1) + 3 \implies 10 = a + b + 3 \implies a + b = 7 \quad (2)
\]

Step 4: Solve the system of equations

We have the system:
\[

$$\begin{cases} a - b = -5 \\ a + b = 7 \end{cases}$$

\]
Add the two equations:
\[
(a - b) + (a + b) = -5 + 7 \implies 2a = 2 \implies a = 1
\]
Substitute \( a = 1 \) into equation (2):
\[
1 + b = 7 \implies b = 6
\]

Step 5: Verify the function

The quadratic function is \( y = x^2 + 6x + 3 \). Let's verify with \( x = -2 \):
\[
y = (-2)^2 + 6(-2) + 3 = 4 - 12 + 3 = -5
\]
Which matches the table. For \( x = 2 \):
\[
y = (2)^2 + 6(2) + 3 = 4 + 12 + 3 = 19
\]
Which also matches the table.

So the relationship between \( x \) and \( y \) is \( \boldsymbol{y = x^2 + 6x + 3} \).