3. \( \left[\left.\begin{array}{l|ccccc}x & -4 & -3 & -2 & -1 & 0 \\ y & -11 & -18 & -21 & -20 & -15\end{array} \right\rvert\,\right. \)

It looks like you have provided a table of values for \(x\) and \(y\). The table appears to represent a set of points that can be used to analyze the relationship between \(x\) and \(y\). The values are as follows: \[ \begin{array}{c|ccccc} x & -4 & -3 & -2 & -1 & 0 \\ \hline y & -11 & -18 & -21 & -20 & -15 \\ \end{array} \] To analyze this data, we can look for patterns, such as whether the relationship between \(x\) and \(y\) is linear, quadratic, or follows some other form. ### Step 1: Calculate the differences in \(y\) Let's calculate the first differences of \(y\): - From \(x = -4\) to \(x = -3\): \(-18 - (-11) = -7\) - From \(x = -3\) to \(x = -2\): \(-21 - (-18) = -3\) - From \(x = -2\) to \(x = -1\): \(-20 - (-21) = 1\) - From \(x = -1\) to \(x = 0\): \(-15 - (-20) = 5\) The first differences are: \(-7, -3, 1, 5\). ### Step 2: Calculate the second differences Now, let's calculate the second differences: - From \(-7\) to \(-3\): \(-3 - (-7) = 4\) - From \(-3\) to \(1\): \(1 - (-3) = 4\) - From \(1\) to \(5\): \(5 - 1 = 4\) The second differences are constant and equal to \(4\), which suggests that the relationship between \(x\) and \(y\) is quadratic. ### Step 3: Formulate a quadratic equation A quadratic equation can be expressed in the form: \[ y = ax^2 + bx + c \] To find the coefficients \(a\), \(b\), and \(c\), we can use the points provided. We can set up a system of equations using three points from the table. Let's use the points \((-4, -11)\), \((-2, -21)\), and \((0, -15)\). 1. For \((-4, -11)\): \[ -11 = a(-4)^2 + b(-4) + c \implies -11 = 16a - 4b + c \] 2. For \((-2, -21)\): \[ -21 = a(-2)^2 + b(-2) + c \implies -21 = 4a - 2b + c \] 3. For \((0, -15)\): \[ -15 = a(0)^2 + b(0) + c \implies -15 = c \] ### Step 4: Solve the system of equations Substituting \(c = -15\) into the first two equations: 1. \( -11 = 16a - 4b - 15 \) \[ 4 = 16a - 4b \implies 4a - b = 1 \quad \text{(Equation 1)} \] 2. \( -21 = 4a - 2b - 15 \) \[ -6 = 4a - 2b \implies 2a - b = -3 \quad \text{(Equation 2)} \] Now we can solve Equations 1 and 2: From Equation 1: \[ b = 4a - 1 \] Substituting into Equation 2: \[ 2a - (4a - 1) = -3 \implies 2a - 4a + 1 = -3 \implies -2a = -4 \implies a = 2 \] Now substituting \(a = 2\) back into Equation 1: \[ 4(2) - b = 1 \implies 8 - b = 1 \implies b = 7 \] ### Final Equation Thus, we have: \[ a = 2, \quad b = 7, \quad c = -15 \] The quadratic equation that fits the data is: \[ y = 2x^2 + 7x - 15