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 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...
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • 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
  • anonymous
are you agree?
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anonymous
  • anonymous
yes, but i thought the value of c caused a vertical shift?

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anonymous
  • 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
  • anonymous
ok so also, the answer says that option 3 is correct as well, can you help with that too?
anonymous
  • anonymous
lets think together..
anonymous
  • anonymous
first point : f(0)=c
anonymous
  • anonymous
right, and i can't figure out where to go from there...
anonymous
  • anonymous
ah! do you know the coordinates of the vertex of the parabola formula: Xv=-(b/2a) Yv=-delta/4a ??
anonymous
  • anonymous
yes i know this one
anonymous
  • anonymous
we have Xv=-1
anonymous
  • anonymous
so b=2a, but "a" is negative...
anonymous
  • anonymous
so "b" is negative too
anonymous
  • anonymous
ok, but how does that affect t?
anonymous
  • anonymous
i think i'm going to sleep on it...i'm tired. thank you for your help with part 1!!
anonymous
  • 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
  • anonymous
i'll make that note for the morning :) good night!
anonymous
  • anonymous
good night!

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