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- 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 y =f(x) is -1. If t is a real number for which f(t) > f(0), which of the following must be true.
1: -2 < t < 0
2: f(t) < f(-2)
3: f(t) > f(1)
really struggling to conceptualize this...

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- anonymous

- jamiebookeater

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- anonymous

n22, the answer 1 must be correct.
The point c is on the right of vertex x=-1.. so values greater then c must be on the "top" of parabola...
im drawing..

- anonymous

are you agree?

- anonymous

yes, but i thought the value of c caused a vertical shift?

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- anonymous

this is a generic chart because c must be any real value.. And you right c causes the vertical shift up or down

- anonymous

ok so also, the answer says that option 3 is correct as well, can you help with that too?

- anonymous

lets think together..

- anonymous

first point : f(0)=c

- anonymous

right, and i can't figure out where to go from there...

- anonymous

ah! do you know the coordinates of the vertex of the parabola formula:
Xv=-(b/2a)
Yv=-delta/4a
??

- anonymous

yes i know this one

- anonymous

we have Xv=-1

- anonymous

so b=2a, but "a" is negative...

- anonymous

so "b" is negative too

- anonymous

ok, but how does that affect t?

- anonymous

i think i'm going to sleep on it...i'm tired. thank you for your help with part 1!!

- anonymous

you're welcome.. but i think the answer 3can be worng... values greater then c must to be in interval on the "top" of parabola..
good look

- anonymous

i'll make that note for the morning :) good night!

- anonymous

good night!

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