**Answer:**

The temperature, in December is 47 °F

**Step-by-step explanation:**

The given data for the temperature are;

Month Temp (°F)

April 59

May 68

June 71

July 78

August 74

Let x represent the month, and f(x) = y represent the temperature, we have;

y = a·x² + b·x + c

When x = 1, we have, f(1) = a·1² + b·1 + c = 59

f(1) = a + b + c = 59...(1)

When x = 2, we have, f(2) = a·2² + b·2 + c = 68

f(2) = 4·a + 2·b + c = 68...(2)

When x = 3, we have, f(3) = a·3² + b·3 + c = 71

f(3) = 9·a + 3·b + c = 71...(3)

When x = 4, we have, f(4) = a·4² + b·4 + c = 78

f(3) = 16·a + 4·b + c = 78...(4)

When x = 5, we have, f(5) = a·5² + b·5 + c = 74

f(3) = 25·a + 5·b + c = 74...(5)

Subtracting equation (3) from equation (5), we get;

25·a + 5·b + c - (9·a + 3·b + c) = 74 - 71

16·a + 2·b = 3...(6)

Subtracting equation (1) from equation (2), we get;

4·a + 2·b + c - (a + b + c) = 68 - 59

3·a + b = 9...(7)

Multiply equation (7) by 2 and substract from equation (6) gives;

16·a + 2·b - 2 × (3·a + b) = 3 - 2 × 9

16·a + 2·b - 6·a - 2·b = 3 - 2 × 9

16·a - 6·a + 2·b - 2·b = 3 - 2 × 9 = -15

10·a = -15

a = -15/10 = -1.5

a = -1.5

From, 3·a + b = 9, we have;

3 × (-1.5) + b = 9

b = 9 + 4.5 =13.5

b = 13.5

From, a + b + c = 59, we have;

-1.5 + 13.5 + c = 59

c = 59 - (-1.5 + 13.5) = 47

c = 47

The quadratic equation becomes, y = a·x² + b·x + c = -1.5·x² + 13.5·x + 47

f(x) = y = -1.5·x² + 13.5·x + 47

For December, we have, x = 9, and f(x) = -1.5×9² + 13.5×9 + 47 = 47

The temperature, in December = 47 °F