the function f is defined for all real numbers x by f(x) = ax^2 + bx + c where a, b and c are constants and a is negative, in the xy-plane, the x coordinate of the vertex of the parabola is y=f(x) is -1. if t is a number for which f(t) > f(0), which of the following (any or all) must be true? 1: -2 < t < 0 2: f(t) < f(-2) 3: f(t) > f(1) please help!

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the function f is defined for all real numbers x by f(x) = ax^2 + bx + c where a, b and c are constants and a is negative, in the xy-plane, the x coordinate of the vertex of the parabola is y=f(x) is -1. if t is a number for which f(t) > f(0), which of the following (any or all) must be true? 1: -2 < t < 0 2: f(t) < f(-2) 3: f(t) > f(1) please help!

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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we know this parabola is oven downward because a is negative and the vertex is (-1,k)
f(t)>f(0) ....
f(t)=a(t+1)^2+k and f(0)=a(0+1)^2+k=a+k and f(t)>f(0) so we have a(t+1)^2+k=a+k

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solve for t
so -2
so are you solving as a system here?
i solve that equation by using remedial algebra
subtract k on both sides
divide by a on both sides
so (t+1)^2 = 1
yes
square root both sides
you get t+1=+ or - 1
t=-1-1,1-1=-2,0
ok, i think i understand that. also, it says option 3 is true, can you help with that?
lets see we gots to think I can't believe I could actually about the first one lol
i have a question where i have equal sign, it should had been a greater than sign a(t+1)^2+k>a+k I don't know if you caught that or not
yes, i assumed it was since the problem is dealing with inequality
okay cool im still thinking about 3
i wonder what happens if we assume f(t)>f(1) and solve for t again im going to try
we get -3
you said the last one is true?
the interval (-3,1) is not included in the first interval we got so i would think the 3rd one would be false unless i made a mistake somewhere
right, but possibly the range works somehow? i have the harder time getting why option 3 works...
oh wait we have t>1 or t<-3 not -3
oomg wait are you for sure number 1 is right? is it true or false?
ok sqrt(x^2)=|x|, so we have sqrt((t+1)^2)>1 and sqrt((t+1)^2))<-1 so we have t>0 or t<-2 and for the last one we got t<-3 or t>1 which are included in that interval so number 3 is true and number 1 is false now what about number 2...
for the second one we get -20 or t<-2 so number 2 is false
i get it. the only true one is number 3 the others are FALSE!
what do you think? any questions/
i'm fixing to go to bed if you dont have any questions
I really don't want to leave if you have questions, but I'm sleepy and I don't know if I should wait or not
i think i kind of get it, although i'm not sure where the absolute value comes in
sqrt(x^2)=|x| just think about it whats sqrt((-3)^2) and is equal to |-3|
since I had something square and i took the square root of it that made me think of the absolute value thing where you change the sign of the number when the inequality sign is flipped
ohh, ok that's right, i think i understand now, that was the one part that i didn't follow
thank you so much for your help!
the first time i did part one i got the interval wrong i don't know if you notice i corrected myself on that one or not
yes, i saw, thank you for checking
we know that interval is true since the question said if it was true right?
so we based everything off that result
exactly, i think we have to assume that base on the problem
yes we do because the question says too
i really like that problem that was totally awesome goodnight!
thanks so much for your help, good night to you too!

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