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Mathematics 13 Online
OpenStudy (anonymous):

A sandbag was thrown downward from a building. The function f(t) = -16t2 - 64t + 80 shows the height f(t), in feet, of the sandbag after t seconds: Part A: Factor the function f(t) and use the factors to interpret the meaning of the x-intercept of the function. (4 points) Part B: Complete the square of the expression for f(x) to determine the vertex of the graph of f(x). Would this be a maximum or minimum on the graph? (4 points) Part C: Use your answer in part B to determine the axis of symmetry for f(x)? (2 points)

OpenStudy (anonymous):

HELP PLEASE!!

OpenStudy (yanasidlinskiy):

\(\huge\cal\color{springgreen}{:)}\)

OpenStudy (anonymous):

can you help me on this one too?

OpenStudy (yanasidlinskiy):

Yes. I lost a bit of connection. My apology:)

OpenStudy (anonymous):

its fine (:

OpenStudy (yanasidlinskiy):

\[f(t)=-16t^2-64t+80\]

OpenStudy (anonymous):

is that part A?

OpenStudy (yanasidlinskiy):

(A) \[-16(t-1)(t+5)\]

OpenStudy (yanasidlinskiy):

That is your original equation

OpenStudy (anonymous):

so the answer for Part A would be the second equation?

OpenStudy (yanasidlinskiy):

So f(t)=0 when t = -5 or t -1 Interpretation "The sandbax hits the ground after 1 second." (B) \[144-16(t+2)^2\] The vertex is at (-144,-2). It is a maximum.

OpenStudy (yanasidlinskiy):

Yes. It ouwld be the second equation. you are correct:)

OpenStudy (anonymous):

okay(:

OpenStudy (yanasidlinskiy):

Sorry about the misspelling. My typing isn't cooperating today..lol..:)

OpenStudy (anonymous):

hahah its fine, would where you put "(B)" be part B?

OpenStudy (yanasidlinskiy):

(C) Using part (B) we know the axis of symmetry is the line x =-2.

OpenStudy (yanasidlinskiy):

Yes:)

OpenStudy (anonymous):

and for Part C would I just answer with "x=-2"?

OpenStudy (yanasidlinskiy):

Yep!!!!!:) Great Job!:)

OpenStudy (anonymous):

Thank you!(: could you help me on 2 more?

OpenStudy (yanasidlinskiy):

Yep! Sure! I have plenty of time:) But close this one and open a new one:)

OpenStudy (anonymous):

okay no problem(:

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