anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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myininaya
  • myininaya
we know this parabola is oven downward because a is negative and the vertex is (-1,k)
myininaya
  • myininaya
f(t)>f(0) ....
myininaya
  • 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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myininaya
  • myininaya
solve for t
myininaya
  • myininaya
so -2
anonymous
  • anonymous
so are you solving as a system here?
myininaya
  • myininaya
i solve that equation by using remedial algebra
myininaya
  • myininaya
subtract k on both sides
myininaya
  • myininaya
divide by a on both sides
anonymous
  • anonymous
so (t+1)^2 = 1
myininaya
  • myininaya
yes
myininaya
  • myininaya
square root both sides
myininaya
  • myininaya
you get t+1=+ or - 1
myininaya
  • myininaya
t=-1-1,1-1=-2,0
anonymous
  • anonymous
ok, i think i understand that. also, it says option 3 is true, can you help with that?
myininaya
  • myininaya
lets see we gots to think I can't believe I could actually about the first one lol
myininaya
  • 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
  • anonymous
yes, i assumed it was since the problem is dealing with inequality
myininaya
  • myininaya
okay cool im still thinking about 3
myininaya
  • myininaya
i wonder what happens if we assume f(t)>f(1) and solve for t again im going to try
myininaya
  • myininaya
we get -3
myininaya
  • myininaya
you said the last one is true?
myininaya
  • 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
  • anonymous
right, but possibly the range works somehow? i have the harder time getting why option 3 works...
myininaya
  • myininaya
oh wait we have t>1 or t<-3 not -3
myininaya
  • myininaya
oomg wait are you for sure number 1 is right? is it true or false?
myininaya
  • 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
  • myininaya
for the second one we get -20 or t<-2 so number 2 is false
myininaya
  • myininaya
i get it. the only true one is number 3 the others are FALSE!
myininaya
  • myininaya
what do you think? any questions/
myininaya
  • myininaya
i'm fixing to go to bed if you dont have any questions
myininaya
  • 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
  • anonymous
i think i kind of get it, although i'm not sure where the absolute value comes in
myininaya
  • myininaya
sqrt(x^2)=|x| just think about it whats sqrt((-3)^2) and is equal to |-3|
myininaya
  • 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
  • anonymous
ohh, ok that's right, i think i understand now, that was the one part that i didn't follow
anonymous
  • anonymous
thank you so much for your help!
myininaya
  • 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
  • anonymous
yes, i saw, thank you for checking
myininaya
  • myininaya
we know that interval is true since the question said if it was true right?
myininaya
  • myininaya
so we based everything off that result
anonymous
  • anonymous
exactly, i think we have to assume that base on the problem
myininaya
  • myininaya
yes we do because the question says too
myininaya
  • myininaya
i really like that problem that was totally awesome goodnight!
anonymous
  • anonymous
thanks so much for your help, good night to you too!

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