Question\begin{align*} &\text{Given the quadratic function:} \\ &y = -2x^2 - 4x + 3 \\ &\text{Calculate the first and second differences, } \Delta y \text{ and } \Delta^2 y, \text{ for the following values of } x: \\ &x = -3, -2, -1, 0, 1, 2, 3 \end{align*}
Studdy Solution
STEP 1
1. The function given is a quadratic function.
2. First differences () are calculated by subtracting consecutive -values.
3. Second differences () are calculated by subtracting consecutive first differences.
STEP 2
1. Calculate -values for each given .
2. Compute the first differences ().
3. Compute the second differences ().
STEP 3
Calculate -values for each given using the function .
\begin{align*}
x = -3, & \quad y = -2(-3)^2 - 4(-3) + 3 = -18 + 12 + 3 = -3 \\
x = -2, & \quad y = -2(-2)^2 - 4(-2) + 3 = -8 + 8 + 3 = 3 \\
x = -1, & \quad y = -2(-1)^2 - 4(-1) + 3 = -2 + 4 + 3 = 5 \\
x = 0, & \quad y = -2(0)^2 - 4(0) + 3 = 3 \\
x = 1, & \quad y = -2(1)^2 - 4(1) + 3 = -2 - 4 + 3 = -3 \\
x = 2, & \quad y = -2(2)^2 - 4(2) + 3 = -8 - 8 + 3 = -13 \\
x = 3, & \quad y = -2(3)^2 - 4(3) + 3 = -18 - 12 + 3 = -27 \\
\end{align*}
STEP 4
Compute the first differences () by subtracting consecutive -values.
\begin{align*}
\Delta y_{-3 \to -2} &= 3 - (-3) = 6 \\
\Delta y_{-2 \to -1} &= 5 - 3 = 2 \\
\Delta y_{-1 \to 0} &= 3 - 5 = -2 \\
\Delta y_{0 \to 1} &= -3 - 3 = -6 \\
\Delta y_{1 \to 2} &= -13 - (-3) = -10 \\
\Delta y_{2 \to 3} &= -27 - (-13) = -14 \\
\end{align*}
STEP 5
Compute the second differences () by subtracting consecutive first differences.
\begin{align*}
\Delta^2 y_{-3 \to -2 \to -1} &= 2 - 6 = -4 \\
\Delta^2 y_{-2 \to -1 \to 0} &= -2 - 2 = -4 \\
\Delta^2 y_{-1 \to 0 \to 1} &= -6 - (-2) = -4 \\
\Delta^2 y_{0 \to 1 \to 2} &= -10 - (-6) = -4 \\
\Delta^2 y_{1 \to 2 \to 3} &= -14 - (-10) = -4 \\
\end{align*}
The first differences are and the second differences are .
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