Sue removes the plug from a trough to drain the water inside. The volume, in gallons, in the trough after it has been unplugged can be modeled by [tex]$4t^2 - 32t + 63$[/tex], where [tex]$t$[/tex] is time, in minutes.

A. Select the correct property that will give Sue the amount of time it takes the trough to drain.
A. minimum
B. maximum
C. [tex][tex]$y$[/tex][/tex]-intercept
D. zero

B. Select the expression that will reveal the property.
A. [tex]$4(0)^2 - 32(0) + 63$[/tex]
B. [tex]$(2t - 7)(2t - 9)$[/tex]
C. [tex][tex]$4(t - 4)^2 - 1$[/tex][/tex]
D. [tex]$4(t - 8)^2 + 47$[/tex]



Answer :

To determine the amount of time it takes to drain the trough completely, let's go through the given details and select the correct property and expression.

A. Select the correct property:
To find the amount of time it takes for the trough to drain, we need to know when the volume of the water becomes zero. This occurs at the "zero" of the volume function, which is when the polynomial [tex]\(4t^2 - 32t + 63\)[/tex] equals zero. Hence, the correct property is:
(D) zero

B. Select the expression that will reveal the property:
To find the zeros of the quadratic function [tex]\(4t^2 - 32t + 63\)[/tex], we can factor the quadratic expression.

When we factor [tex]\(4t^2 - 32t + 63\)[/tex], we get:
[tex]\[4t^2 - 32t + 63 = (2t - 7)(2t - 9)\][/tex]

So the expression that reveals the zeros of the quadratic function is:
(B) [tex]\((2t - 7)(2t - 9)\)[/tex]

This expression shows us that when either [tex]\(2t - 7 = 0\)[/tex] or [tex]\(2t - 9 = 0\)[/tex], the volume is zero.

By solving these equations:
[tex]\[2t - 7 = 0\][/tex]
[tex]\[t = \frac{7}{2}\][/tex]

[tex]\[2t - 9 = 0\][/tex]
[tex]\[t = \frac{9}{2}\][/tex]

Therefore, the zeros of the function [tex]\(4t^2 - 32t + 63\)[/tex] are at [tex]\(t = \frac{7}{2}\)[/tex] and [tex]\(t = \frac{9}{2}\)[/tex] minutes.

In conclusion:
A. (D) zero
B. (B) [tex]\((2t - 7)(2t - 9)\)[/tex]