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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

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

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?

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?

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 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|

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!

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!