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!

- anonymous

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- myininaya

we know this parabola is oven downward because a is negative and the vertex is (-1,k)

- myininaya

f(t)>f(0) ....

- myininaya

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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## More answers

- myininaya

solve for t

- myininaya

so -2

- anonymous

so are you solving as a system here?

- myininaya

i solve that equation by using remedial algebra

- myininaya

subtract k on both sides

- myininaya

divide by a on both sides

- anonymous

so (t+1)^2 = 1

- myininaya

yes

- myininaya

square root both sides

- myininaya

you get t+1=+ or - 1

- myininaya

t=-1-1,1-1=-2,0

- anonymous

ok, i think i understand that. also, it says option 3 is true, can you help with that?

- myininaya

lets see we gots to think I can't believe I could actually about the first one lol

- myininaya

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

- anonymous

yes, i assumed it was since the problem is dealing with inequality

- myininaya

okay cool im still thinking about 3

- myininaya

i wonder what happens if we assume f(t)>f(1) and solve for t again im going to try

- myininaya

we get -3

- myininaya

you said the last one is true?

- myininaya

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

- anonymous

right, but possibly the range works somehow?
i have the harder time getting why option 3 works...

- myininaya

oh wait we have t>1 or t<-3 not -3

- myininaya

oomg wait are you for sure number 1 is right? is it true or false?

- myininaya

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...

- myininaya

for the second one we get -20 or t<-2 so number 2 is false

- myininaya

i get it. the only true one is number 3 the others are FALSE!

- myininaya

what do you think? any questions/

- myininaya

i'm fixing to go to bed if you dont have any questions

- myininaya

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

- anonymous

i think i kind of get it, although i'm not sure where the absolute value comes in

- myininaya

sqrt(x^2)=|x| just think about it whats sqrt((-3)^2) and is equal to |-3|

- myininaya

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

- anonymous

ohh, ok that's right, i think i understand now, that was the one part that i didn't follow

- anonymous

thank you so much for your help!

- myininaya

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

- anonymous

yes, i saw, thank you for checking

- myininaya

we know that interval is true since the question said if it was true right?

- myininaya

so we based everything off that result

- anonymous

exactly, i think we have to assume that base on the problem

- myininaya

yes we do because the question says too

- myininaya

i really like that problem that was totally awesome goodnight!

- anonymous

thanks so much for your help, good night to you too!

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