(y 3 - 4y 2 + 7y - 6) (y - 2)

1 answer:

8 0

Y^4-4y^3+7y^2-6y-2y^3+8y^2-14y+12=
y^4-6y^3+15y^2-20y+12

or

y^3-4y^2+7y-6=y^3-4y^2+4y+3y-6=
y(y^2-4y+4)+3(y-2)=
y(y-2)^2+3(y-2)=(y-2)[y(y-2)+3]=(y-2)(y^2-2y+3)

(y^3-4y^2+7y-6)(y-2)=(y-2)^2(y^2-2y+3)

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As a rule, these ballistic motion equations are only applicable for heights above the final resting place. Usually, that is the ground (h=0). Time is always measured as a non-negative quantity beginning at the time of launch.

The function is meaningless after the time the object comes to rest on the ground, so the domain of the function extends from 0 until that time.

These considerations allow us to evaluate the offered statements.

A. The domain never includes negative values. (A is not a true statement)

B. The heights are the same at times that are symmetrical around the time of the maximum height. Here, that time is about 1.9 seconds, not 1.5 seconds, so the heights will be different at 1 and 2 seconds after launch. (B is not a true statement)

C. The graph of the function shows the height reaches 0 at about 4.06 seconds, a little after 4 seconds. (C is true)

D. Based on our answer to C, we know t=10 is outside the domain of the function. (D is not a true statement)

E. The value of the function is 20 when t=0, so the function tells us the orange is 20 feet above the ground when it is launched. (E is true)

The true statements are C and E.