Question

6.3.3 quiz: nonlinear models
what is the quadratic regression equation that fits these data?
a. $y = 2.38x^{2}+0.90x + 9.90$
b. $y = 21.94cdot1.02^{x}$
c. $y = 1.62x^{2}+2.30x + 9.20$
d. $y = 0.34x^{2}+3.40x + 8.60$

Explanation:

Step1: Substitute \(x = 0\) into the equations

For a quadratic equation \(y=ax^{2}+bx + c\), when \(x = 0\), \(y=c\).
Looking at the data, when \(x = 0\), \(y = 8\).

• For option A: When \(x = 0\), \(y=2.38\times0^{2}+0.90\times0 + 9.90=9.90

eq8\)

• For option B: This is an exponential function (\(y = ab^{x}\)), not a quadratic function (\(y=ax^{2}+bx + c\))
• For option C: When \(x = 0\), \(y=1.62\times0^{2}+2.30\times0+9.20 = 9.20

eq8\)

• For option D: When \(x = 0\), \(y=0.34\times0^{2}+3.40\times0 + 8.60=8.60\approx8\) (due to regression approximation)

Step2: Check symmetry (optional, for quadratic functions \(y = ax^{2}+bx + c\), the graph is symmetric about \(x=-\frac{b}{2a}\))

The data has a kind of symmetry around \(x = 0\). For a quadratic function \(y=ax^{2}+bx + c\), the axis of symmetry is \(x =-\frac{b}{2a}\).
For option D: \(a = 0.34\), \(b = 3.40\), \(x=-\frac{3.40}{2\times0.34}=-\frac{3.40}{0.68}=- 5\) (not a good - fit for the symmetry in the data). But since option B is not quadratic and options A and C fail the \(x = 0\) test more severely.

Another way:
We can use the general form of quadratic regression \(y=ax^{2}+bx + c\).
We can also substitute other values (e.g., \(x = 1\)):

• Option A: When \(x = 1\), \(y=2.38\times1^{2}+0.90\times1+9.90=2.38 + 0.90+9.90=13.18\) (actual \(y = 13\))
• Option C: When \(x = 1\), \(y=1.62\times1^{2}+2.30\times1+9.20=1.62+2.30 + 9.20=13.12\) (actual \(y = 13\))
• Option D: When \(x = 1\), \(y=0.34\times1^{2}+3.40\times1+8.60=0.34+3.40+8.60=12.34\) (actual \(y = 13\))

But the \(x = 0\) test is a quick - check for quadratic regression equations of the form \(y=ax^{2}+bx + c\)

Answer:

A. \(y = 2.38x^{2}+0.90x + 9.90\)