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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.
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| 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.