Question

question 1 what is the quadratic regression equation for the data set?

y	x
110	3
50	10
90	3
120	5
30	15
70	9

y = - 0.175x^2 - 3.786x + 121.119
y = - 0.175(3.786)x
y = - 0.175x - 3.786
y = - 0.175x^2 + 3.786x + 121.119

Explanation:

Step1: Recall quadratic regression formula

The general form of a quadratic regression equation is $y = ax^{2}+bx + c$. To find the values of $a$, $b$, and $c$ for a given data - set, we can use statistical software or a calculator with regression capabilities. However, if we assume we are not using external tools, we can use the least - squares method which involves setting up a system of normal equations based on minimizing the sum of the squared residuals $\sum_{i = 1}^{n}(y_{i}-(ax_{i}^{2}+bx_{i}+c))^{2}$. But in this case, we can check each option by substituting the $x$ - values from the data set into the equations and seeing which one gives the closest $y$ - values.
Let's take the first option $y=-0.175x^{2}-3.786x + 121.119$.
For $x = 6$:
$y=-0.175\times6^{2}-3.786\times6 + 121.119$
$y=-0.175\times36-22.716 + 121.119$
$y=-6.3-22.716 + 121.119$
$y=92.103$ (close to 100)
For $x = 3$:
$y=-0.175\times3^{2}-3.786\times3 + 121.119$
$y=-0.175\times9-11.358 + 121.119$
$y=-1.575-11.358 + 121.119$
$y=108.186$ (close to 110)
For $x = 10$:
$y=-0.175\times10^{2}-3.786\times10 + 121.119$
$y=-17.5-37.86 + 121.119$
$y=65.759$ (not so close to 50)

Let's check the fourth option $y=-0.175x^{2}+3.786x + 121.119$
For $x = 6$:
$y=-0.175\times6^{2}+3.786\times6 + 121.119$
$y=-0.175\times36 + 22.716+121.119$
$y=-6.3 + 22.716+121.119$
$y=137.535$ (not close to 100)

We can rule out the second option $y=-0.175(3.786)x$ which is a linear - like form (not a quadratic in the correct $ax^{2}+bx + c$ form). The third option $y=-0.175x-3.786$ is also linear.

By substituting more points from the data set into the first option $y=-0.175x^{2}-3.786x + 121.119$, we find that it gives values that are relatively close to the data - set values.

Answer:

$y=-0.175x^{2}-3.786x + 121.119$