Questions: Given the points graphed in the following figure, use quadratic regression to find the quadratic function of best fit. Round the coefficients to three decimal places, if necessary.

Solution Steps

From the graph, identify the coordinates of the points. The points appear to be approximately:

• (-6, -2)
• (-3, 2)
• (0, 4)
• (3, 2)

The general form of a quadratic equation is \( y = ax^2 + bx + c \). We need to find the coefficients \( a \), \( b \), and \( c \) that best fit the given points.

Substitute each point into the quadratic equation to form a system of equations:

1. For (-6, -2): \(-2 = a(-6)^2 + b(-6) + c\)
2. For (-3, 2): \(2 = a(-3)^2 + b(-3) + c\)
3. For (0, 4): \(4 = a(0)^2 + b(0) + c\)
4. For (3, 2): \(2 = a(3)^2 + b(3) + c\)

Solve the system of equations to find the values of \( a \), \( b \), and \( c \).

1. \(-2 = 36a - 6b + c\)
2. \(2 = 9a - 3b + c\)
3. \(4 = c\)
4. \(2 = 9a + 3b + c\)

From the third equation, \( c = 4 \).

Substitute \( c = 4 \) into the other equations:

1. \(-2 = 36a - 6b + 4\) ⟹ \(-6 = 36a - 6b\) ⟹ \(-1 = 6a - b\)
2. \(2 = 9a - 3b + 4\) ⟹ \(-2 = 9a - 3b\) ⟹ \(-2 = 3a - b\)
3. \(2 = 9a + 3b + 4\) ⟹ \(-2 = 9a + 3b\)

Now solve the simplified system:

1. \(-1 = 6a - b\)
2. \(-2 = 3a - b\)

Subtract the second equation from the first: \(-1 - (-2) = (6a - b) - (3a - b)\) \(1 = 3a\) \(a = \frac{1}{3}\)

Substitute \( a = \frac{1}{3} \) into \(-2 = 3a - b\): \(-2 = 3(\frac{1}{3}) - b\) \(-2 = 1 - b\) \(-3 = -b\) \(b = 3\)

Final Answer