determining the regression equation\nthe table below shows the approximate height of a ball thrown up in the…

determining the regression equation\nthe table below shows the approximate height of a ball thrown up in the air after x seconds. which quadratic model best represents the data?\n| time (seconds) | height (feet) |\n| ---- | ---- |\n| 0 | 5 |\n| 1 | 90 |\n| 2 | 140 |\n| 3 | 160 |\n| 4 | 150 |\n○ (f(x)=-16x^{2}+99x + 6)\n○ (f(x)=-36x^{2}+37x + 5)\n○ (f(x)=36x^{2}+37x + 5)\n○ (f(x)=16x^{2}+99x + 6)

determining the regression equation\nthe table below shows the approximate height of a ball thrown up in the air after x seconds. which quadratic model best represents the data?\n| time (seconds) | height (feet) |\n| ---- | ---- |\n| 0 | 5 |\n| 1 | 90 |\n| 2 | 140 |\n| 3 | 160 |\n| 4 | 150 |\n○ (f(x)=-16x^{2}+99x + 6)\n○ (f(x)=-36x^{2}+37x + 5)\n○ (f(x)=36x^{2}+37x + 5)\n○ (f(x)=16x^{2}+99x + 6)

Answer

Explanation:

Step1: Substitute x = 0 into each function

When (x = 0), for (f(x)=-16x^{2}+99x + 6), (f(0)=-16\times0^{2}+99\times0 + 6=6); for (f(x)=-36x^{2}+37x + 5), (f(0)=-36\times0^{2}+37\times0 + 5 = 5); for (f(x)=36x^{2}+37x + 5), (f(0)=36\times0^{2}+37\times0 + 5=5); for (f(x)=16x^{2}+99x + 6), (f(0)=16\times0^{2}+99\times0 + 6=6). The table has height 5 when (x = 0), so we can eliminate the first and fourth - functions.

Step2: Substitute x = 1 into the remaining functions

For (f(x)=-36x^{2}+37x + 5), (f(1)=-36\times1^{2}+37\times1 + 5=-36 + 37+5=6). For (f(x)=36x^{2}+37x + 5), (f(1)=36\times1^{2}+37\times1 + 5=36 + 37+5=78). This doesn't match 90 in the table. Let's use another approach. We know that the general form of a quadratic regression is (y = ax^{2}+bx + c). We can use the least - squares method or substitute the points from the table into the equations. Substitute ((x = 0,y = 5)) into (y=ax^{2}+bx + c), we get (c = 5). Substitute ((x = 1,y = 90)) and (c = 5) into (y=ax^{2}+bx + c), we have (a + b+5 = 90), so (a + b=85). Substitute ((x = 2,y = 140)) and (c = 5) into (y=ax^{2}+bx + c), we get (4a + 2b+5 = 140), so (4a + 2b=135). From (a + b=85), we have (b = 85 - a). Substitute (b = 85 - a) into (4a + 2b=135): [ \begin{align*} 4a+2(85 - a)&=135\ 4a + 170-2a&=135\ 2a&=135 - 170\ 2a&=-35\ a&=-17.5\approx - 16 \end{align*} ] If (a=-16) and (a + b=85), then (b=85+16 = 99) and (c = 6) (a bit off from the first - step analysis but a closer fit overall). Substitute the points ((x,y)) from the table into each equation: For (f(x)=-16x^{2}+99x + 6): When (x = 0), (f(0)=-16\times0^{2}+99\times0 + 6 = 6) (slightly off from 5 in table). When (x = 1), (f(1)=-16\times1^{2}+99\times1 + 6=-16 + 99+6=89\approx90). When (x = 2), (f(2)=-16\times2^{2}+99\times2 + 6=-64 + 198+6=140). When (x = 3), (f(3)=-16\times3^{2}+99\times3 + 6=-144+297 + 6=159\approx160). When (x = 4), (f(4)=-16\times4^{2}+99\times4 + 6=-256+396+6=146\approx150).

Answer:

A. (f(x)=-16x^{2}+99x + 6)