Question

question 3 of 10 what is the quadratic regression equation that fits these data? x y -4 35 -3 20 -2 12 -1 6 0 2 1 6 2 10 3 24 4 38 a. y = 0.84x^2 + 0.22x + 4.38 b. y = 2.09x^2 + 0.33x + 3.06 c. y = 12.09 * 1.01^x d. y = 2.84x^2 + 0.68x + 1.12

Explanation:

Step1: Recall quadratic regression form

The general form of a quadratic regression equation is $y = ax^{2}+bx + c$. We can use a graphing - calculator or statistical software to find the values of $a$, $b$, and $c$ for the given data points $(x,y)$. Another way is to use the least - squares method to solve a system of linear equations. However, a quick check is to substitute the $x$ values from the table into the given equations and see which one gives the closest $y$ values.
Let's start with $x = 0$. When $x = 0$, the equation $y=ax^{2}+bx + c$ simplifies to $y = c$. From the table, when $x = 0$, $y = 2$.

Step2: Check each option for $x = 0$

• Option A: When $x = 0$, $y=0.84\times0^{2}+0.22\times0 + 4.38=4.38$.
• Option B: When $x = 0$, $y=2.09\times0^{2}+0.33\times0 + 3.06 = 3.06$.
• Option C: This is an exponential function $y = 12.09\times1.01^{x}$, not a quadratic function, so we can eliminate it.
• Option D: When $x = 0$, $y=2.84\times0^{2}+0.68\times0+1.12 = 1.12$.

Since when $x = 0$, $y = 2$ from the table, none of the options match exactly at $x = 0$. Let's substitute another point, say $x = 1$.
For the table, when $x = 1$, $y = 6$.

• Option A: When $x = 1$, $y=0.84\times1^{2}+0.22\times1+4.38=0.84 + 0.22+4.38=5.44$.
• Option B: When $x = 1$, $y=2.09\times1^{2}+0.33\times1+3.06=2.09+0.33 + 3.06=5.48$.
• Option D: When $x = 1$, $y=2.84\times1^{2}+0.68\times1+1.12=2.84+0.68 + 1.12=4.64$.

Let's use a more accurate method. We can use a graphing calculator's quadratic regression feature.
Entering the data points $(-4,35),(-3,20),(-2,12),(-1,6),(0,2),(1,6),(2,10),(3,24),(4,38)$ into a graphing calculator and performing quadratic regression, we get $y = 2.09x^{2}+0.33x + 3.06$.

Answer:

B. $y = 2.09x^{2}+0.33x + 3.06$