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Mathematics 9 Online
OpenStudy (anonymous):

A basketball is thrown upwards. The height f(t), in feet, of the basketball at different times t, in seconds, is shown by the following function: f(t) = -16t2 + 16t + 32 Which of the following is a reasonable domain of the graph of the function when the basketball falls from its highest height to the ground?

OpenStudy (anonymous):

these are the choices: 0.5 ≤ x ≤ 1 0.5 ≤ x ≤ 2 1 ≤ x ≤ 2 1 ≤ x ≤ 1.5

OpenStudy (anonymous):

@iambatman @sleepyhead314 @shamim @Loser66 @jim_thompson5910 someone please help!!!

OpenStudy (sleepyhead314):

highest point in this case should be the vertex and the rightest zero x value of vertex = -b/2a given ax^2 + bx + c

OpenStudy (sleepyhead314):

plug in and that'll be the first point... the other point would be solve -16x^2 + 16x + 32 = 0 and take the positive answer, that'll be your other point

OpenStudy (anonymous):

okay one sec

OpenStudy (anonymous):

okay the vertex is (0.5,36) ?

OpenStudy (anonymous):

@sleepyhead314

OpenStudy (sleepyhead314):

um yeah, but you just need the x values to get domain

OpenStudy (anonymous):

....oh

OpenStudy (sleepyhead314):

it's alright, you just did more work than you needed :P no harm done now solve -16x^2 + 16x + 32 = 0 and take the positive answer

OpenStudy (anonymous):

how do i solve this? @sleepyhead314 what method do i use?

OpenStudy (sleepyhead314):

first factor out the greatest common factor (-16) then factor... do you know how to do that? :3

OpenStudy (anonymous):

i think so.. one moment (:

OpenStudy (anonymous):

um i think i may ave used the wrong factoring method.. again.. i got 0=-16(t+0.5)+6

OpenStudy (sleepyhead314):

mmm that looks like vertex form :/ not quite I'm trying to get at something like (x+3)(x-2) = 0 -> x = -3, 2

OpenStudy (anonymous):

Thanks! i definitely used vertex form, oops. so it is the second one right? (B)?

OpenStudy (sleepyhead314):

Correct! :)

OpenStudy (anonymous):

Thank you so much!!! cx

OpenStudy (sleepyhead314):

I'm glad I could help ^_^

OpenStudy (anonymous):

i have 2 more problems, think you could help me with them?

OpenStudy (sleepyhead314):

It's a bit past my bed time, but I can try :)

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