Question

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!

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

solve for t

Answer 5

so -2<t<0

Answer 6

so are you solving as a system here?

Answer 7

i solve that equation by using remedial algebra

Answer 8

subtract k on both sides

Answer 9

divide by a on both sides

Answer 10

so (t+1)^2 = 1

Answer 11

yes

Answer 12

square root both sides

Answer 13

you get t+1=+ or - 1

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

okay cool im still thinking about 3

Answer 20

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

Answer 21

we get -3<t<1

Answer 22

you said the last one is true?

Answer 23

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

Answer 24

right, but possibly the range works somehow?

i have the harder time getting why option 3 works...

Answer 25

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

Answer 26

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

Answer 27

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...

Answer 28

for the second one we get -2<t<0 but we said t>0 or t<-2 so number 2 is false

Answer 29

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

Answer 30

what do you think? any questions/

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

thank you so much for your help!

Answer 38

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

Answer 39

yes, i saw, thank you for checking

Answer 40

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

Answer 41

so we based everything off that result

Answer 42

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

Answer 43

yes we do because the question says too

Answer 44

i really like that problem that was totally awesome goodnight!

Answer 45

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