Question 1203271: Appreciate if you could explain a step by step in detail how to get the answer (C).
Tried to use these Y = a(x-h)² + k and -b/2a but not sure this is the way to do so.
Thank you!
The function f is defined for all real numbers x by f(x)=ax^2+bx+c where a, b, 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 number for which f(t)>f(0), which of the following must be true?
I. -2 < t < 0
II. f(t) < f (-2)
III.f(t) > f (1)
(A) I only
(B) II only
(C) I and III only
(D) II and III only
(E) I, II, and III
Found 3 solutions by ikleyn, math_tutor2020, greenestamps:
Answer by ikleyn(52775)   (Show Source):
You can put this solution on YOUR website! .
The function f is defined for all real numbers x by f(x)=ax^2+bx+c where a, b, 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 number for which f(t) > f(0), which of the following must be true?
I. -2 < t < 0
II. f(t) < f (-2)
III. f(t) > f (1)
(A) I only
(B) II only
(C) I and III only
(D) II and III only
(E) I, II, and III
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This problem is not to solve it by applying formal algebraic transformations/inequalities.
It is for MENTAL solution, using reasoning.
You should apply your geometric imagination.
The preliminary analysis
The given parabola is opened downward and has the vertex at x= -1 (given).
It means that the parabola has the maximum at x= -1.
On the coordinate plane, draw (mentally) such a parabola and horizontal line y = f(0).
It is clear, that this line is BELOW the vertex.
The point (0,f(0)) lies on the parabola and on the horizontal line.
Hence, the symmetric point (-2, f(-2)) ALSO lies on the parabola and on the same horizontal line.
So, you conclude that at -2 < t < 0, the points of parabola are ABOVE the points of the line.
Geometrically, the opposite is also OBVIOUS: if the point of the line is under the parabola, then -2 < t < 0.
Also, notice that f(0) = f(-2) due to symmetry relative the axis x= -1.
Now we are ready to analyze cases (I), (II) and (III).
So, if t is a number for which f(t) > f(0), it means that the parabola is above the line,
and hence, (I) is TRUE: -2 < t < 0.
Next, if t is a number for which f(t) > f(0), it means that the parabola is above the line,
and hence, (II) is NOT NOT.
And finally, if t is a number for which f(t) > f(0), it means that f(t) > f(0) > f(1),
i.e. (III) is TRUE.
+--------------------------------------------------+
| At this point, the major part of the problem |
| is just done. |
+--------------------------------------------------+
Having it, the rest is just  :
of 5 options (A), (B), (C), (D) and (E), only (C) is true.
Solved.
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This problem seems to be complicated ONLY BECAUSE
they suppressed you by great number of words and options.
Answer by math_tutor2020(3816)  (Show Source):
You can put this solution on YOUR website!
Hint:
Let a = -1 and consider f(x) = -1(x+1)^2
The parabola f(x) = -1(x+1)^2 is in vertex form y = a(x-h)^2+k with vertex (h,k) = (-1,0)
For this example f(0) = -1 and f(-1) = 0
f(-1) is the largest output possible because it's at the vertex peak. This parabola opens downward.
Answer by greenestamps(13198)  (Show Source):
You can put this solution on YOUR website!
The conditions tell us the parabola opens downward, with axis of symmetry at x=-1.
We are given that f(t) is greater than f(0).
By symmetry, f(-2) = f(0) (-2 and 0 are the same distance from the axis of symmetry).
So f(-2) = f(0), f(t) is greater than f(0), the parabola opens downward, and the axis of symmetry is at x=-1. That means that t must be between -2 and 0, which is condition I.
So I is true.
Given that f(t) is greater than f(0) and f(-2) = f(0), f(t) is also greater than f(-2). So condition II is false.
We are given that f(t) is greater than f(0). x=1 is farther to the right of the axis of symmetry than x=0; since the parabola opens downward, f(t) is greater than f(1).
So III is true.
I and III are true and II is false, so the answer is (C) I and III only.
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