Question

Use technology to find the quadratic regression curve through the given points. HINT [See Example 5.] (Round all coefficients to four decimal places.) (1, 4), (3, 6), (4, 5), (5, 2)

y(x) =

Answer 1

The coefficients for the quadratic regression curve are

a = (-2/3) = -0.6667

b = (11/3) = 3.6667

c = 1 = 1.0000

y(x) = -0.6667x² + 3.6667x + 1.0000

Explanation:

Quadratic regression curve gives a general expression of

y = ax² + bx + c

And the points on the curve include

(1, 4), (3, 6), (4, 5), (5, 2)

Taking the points one at a time and substituting them into general quadratic curve expression

(1, 4), x = 1, y = 4

y = ax² + bx + c

4 = a + b + c (eqn 1)

(3, 6), x = 3, y = 6

6 = a(3²) + b(3) + c

6 = 9a + 3b + c (eqn 2)

(4, 5), x = 4, y = 5

5 = a(4²) + b(4) + c

5 = 16a + 4b + c (eqn 3)

Combining the 3 equations and solving simultaneously

4 = a + b + c

6 = 9a + 3b + c

5 = 16a + 4b + c

From eqn 1, c = 4 - a - b

Substituting this into eqn 2 and 3, we have

6 = 9a + 3b + 4 - a - b

2 = 8a + 2b (*)

5 = 16a + 4b + 4 - a - b

1 = 15a + 3b (**)

8a + 2b = 2

15a + 3b = 1

a = (-2/3), b = (11/3)

c = 4 - a - b

c = 4 - (-2/3) - (11/3)

c = 1

Hence, the coefficients for the quadratic regression curve are

a = (-2/3) = -0.6667

b = (11/3) = 3.6667

c = 1 = 1.0000

y(x) = -0.6667x² + 3.6667x + 1.0000

Hope this Helps!!!