The table below shows the height of a ball x seconds after being kicked.

What values, rounded to the nearest whole number, complete the quadratic regression equation that models the data?
f(x) = __ x^2 + _ x + 0
Based on the regression equation and rounded to the nearest whole number, what is the estimated height after 0.25 seconds?

__ feet

Part A:

The general form of a quadratic function is given by:

[tex]y=ax^2+bx+c[/tex]

Given that the table shows the height (y) of a ball x seconds after being kicked.
Thus, the time represents the x-values while the height represents the y-values.

From the table it can be seen that when the time is 0, the height is also 0, thus:

[tex]0=a(0)^2+b(0)+c \\ \\ \Rightarrow c=0 \ . \ . \ . \ (1)[/tex]

Also from the table it can be seen that when the time is 1, the height is 65, thus:

[tex]65=a(1)^2+b(1)+0 \\ \\ \Rightarrow a+b=65 \ . \ . \ . \ (2)[/tex]

Also from the table it can be seen that when the time is 2, the height is 95, thus:

[tex]95=a(2)^2+b(2)+0 \\ \\ \Rightarrow 4a+2b=95 \ . \ . \ . \ (3)[/tex]

Multiplying equation (2) by 2, we have:
2a + 2b = 130 . . . (4)

Subtracting equation (4) from equation (3) gives:
2a = -35
or a = -35 / 2 = -17.5

Substituting for a into equation (2) gives:
-17.5 + b = 65
or b = 65 + 17.5 = 82.5

Therefore, the quadratic regression equation that models the data is given by f(x) = -18x^2 +83x + 0

Part B:

Using the function obtained in part a, with x = 0.25, we have

[tex]f(x) = -18(0.25)^2 +83(0.25) + 0 \\ \\ =-18(0.0625)+20.75=-1.125+20.75 \\ \\ =19.625[/tex]

Therefore, the estimated height after 0.25 seconds is 20 feet.

Аноним Аноним · Answer 2 · 2017-11-18T14:31:31+01:00

A quadratic regression equation is of the form
y = c₃x² + c₂x +c₁

Coefficients c₁,c₂,c₃ are determined by minimizing the least squared error
[tex]S = \sum _{i=1}^{n} [y_{i} - (c_{3}x_{i}^2+c_{2}x_{i}+c_{1})]^{2}[/tex]
This means that
[tex] \frac{\partial S}{\partial c_{1}} = \frac{\partial S}{\partial c_{2}} = \frac{\partial S}{\partial c_{3}} =0[/tex]

This creates the matrix equation for determining the c-coefficients:
[tex]\begin{bmatrix} n&\sum x&\sum x^{2}\\\sum x&\sum x^{2}&\sum x^{3}\\\sum x^{2}&\sum x^{3}&\sum x^{4} \end{bmatrix} \begin{bmatrix} c_{1}\\c_{2}\\c_{3}\end{bmatrix} = \begin{bmatrix} \sum y\\\sum xy\\ \sum x^{2}y\end{bmatrix}[/tex]

For the given problem,
the matrix equation is
[tex]\begin{bmatrix} 7&10.5&22.75\\10.5&22.75&55.125\\22.75&55.125&142.1875 \end{bmatrix} \begin{bmatrix} c_{1}\\c_{2}\\c_{3}\end{bmatrix} = \begin{bmatrix}475\\935\\2125 \end{bmatrix} [/tex]

The solution for the coefficients is
c₁ = -0.3571
c₂ = 81.0714
c₃ = -16.4286

The graph of the fitted data is shown below.

When x = 0.25,
y = c₃(0.25)² + c₂(0.25) + c₁ = 18.884

The regression equation is
f(x) = -16.4286x² + 81.0714x - 0.3571
When rounded to the nearest whole number,
f(x) = -16x² + 81x + 0
and
f(0.25) = 19.25

Answers:
f(x) = -16x² + 81x + 0 (approximated)
f(0.25) = 19.25 (approximated)