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Answer to this question cannot be D, it should be B (II only).
To solve this question you don't need to calculate a, b, and c. We should put the given three points on the XY-plane and we'll get: parabola \(f(x)=ax^2+bx+c\) will have the vertex at \(x=1\), halfway between \(x=-3\) and \(x=5\), as \(f(-3)=0=f(5)\). Plus, as \(f(0)=3\) the parabola would be downward. As the parabola is downward, value of f(x) at x=1, f(1), is the the greatest value of f(x). As we move from x=1 to either of direction the value of f(x) will decrease. So any \(f(m)\) would be greater than \(f(n)\) if \(m\) is closer to \(x=1\) than \(n\). For example \(f(9)>f(10)>f(-12)\) or \(f(6)=f(-4)\) or \(f(-5)=f(7)>f(-100)\).
Hence I. \(f(-1)>f(2)\) is false, as \(x=2\) is 1 farther from \(x=1\) and \(x=-1\) is 2 farther than \(x=1\). (True inequality would be \(f(-1)<f(2)\)).
II. \(f(1)>f(0)\) is true as \(f(1)\) is vertex and more than any \(f(x)\).
III. \(f(2)>f(1)\) is false as no \(f(x)\) is more than \(f(1)\). (The correct would be \(f(2)<f(1)\)).
Re: If the parabola represented by f(x) = ax^2 + bx + c passes [#permalink]
02 Mar 2010, 23:18
Expert's post
1
This post was BOOKMARKED
apoorvasrivastva wrote:
@bunnuel i have dited my post for the st I ..it was a typo error i am sorry abt that
it is f(-1) > f(-2)
i am not clear as to how did u get the vertex as x=1
Intersection points of parabola with x-axis are \((-3,0)\) and \((5,0)\):
-----(-3)--0----(5)---, as parabola is symmetric, the x coordinate of the vertex must be halfway between \(x=-3\) and \(x=5\) --> \(x=\frac{-3+5}{2}=1\). As parabola is downward, \(f(x)\) naturally will have it's max values at vertex \(x=1\), \(f(1)\).
As for the typo in the stem. If I is saying: \(f(-1) > f(-2)\), then it's true as \(x=-1\) is closer to \(x=1\), than \(x=-2\), which means that \(f(-1)>f(-2)\). _________________
Re: If the parabola represented by f(x) = ax^2 + bx + c passes [#permalink]
04 Mar 2010, 20:22
amazing explanation Bunuel. I would normally have found a, b and c and then solved. Thanks for showing me the way to think more than the way you solved the problem.
Re: If the parabola represented by f(x) = ax^2 + bx + c passes [#permalink]
23 Dec 2013, 20:20
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25 Mar 2015, 09:07
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Answer to this question cannot be D, it should be B (II only).
To solve this question you don't need to calculate a, b, and c. We should put the given three points on the XY-plane and we'll get: parabola \(f(x)=ax^2+bx+c\) will have the vertex at \(x=1\), halfway between \(x=-3\) and \(x=5\), as \(f(-3)=0=f(5)\). Plus, as \(f(0)=3\) the parabola would be downward. As the parabola is downward, value of f(x) at x=1, f(1), is the the greatest value of f(x). As we move from x=1 to either of direction the value of f(x) will decrease. So any \(f(m)\) would be greater than \(f(n)\) if \(m\) is closer to \(x=1\) than \(n\). For example \(f(9)>f(10)>f(-12)\) or \(f(6)=f(-4)\) or \(f(-5)=f(7)>f(-100)\).
Hence I. \(f(-1)>f(2)\) is false, as \(x=2\) is 1 farther from \(x=1\) and \(x=-1\) is 2 farther than \(x=1\). (True inequality would be \(f(-1)<f(2)\)).
II. \(f(1)>f(0)\) is true as \(f(1)\) is vertex and more than any \(f(x)\).
III. \(f(2)>f(1)\) is false as no \(f(x)\) is more than \(f(1)\). (The correct would be \(f(2)<f(1)\)).
Hope it's clear.
HI Bunuel,
Please explain the red part , what i understood from making graph , if we move left from the axis point certainly value of f(x) decreases but if we move right side from the axis where x =1, value of f(x) will increase till certain point and then will decrease, so how can we deduce that value of f(x) is max at x =1
Answer to this question cannot be D, it should be B (II only).
To solve this question you don't need to calculate a, b, and c. We should put the given three points on the XY-plane and we'll get: parabola \(f(x)=ax^2+bx+c\) will have the vertex at \(x=1\), halfway between \(x=-3\) and \(x=5\), as \(f(-3)=0=f(5)\). Plus, as \(f(0)=3\) the parabola would be downward. As the parabola is downward, value of f(x) at x=1, f(1), is the the greatest value of f(x). As we move from x=1 to either of direction the value of f(x) will decrease. So any \(f(m)\) would be greater than \(f(n)\) if \(m\) is closer to \(x=1\) than \(n\). For example \(f(9)>f(10)>f(-12)\) or \(f(6)=f(-4)\) or \(f(-5)=f(7)>f(-100)\).
Hence I. \(f(-1)>f(2)\) is false, as \(x=2\) is 1 farther from \(x=1\) and \(x=-1\) is 2 farther than \(x=1\). (True inequality would be \(f(-1)<f(2)\)).
II. \(f(1)>f(0)\) is true as \(f(1)\) is vertex and more than any \(f(x)\).
III. \(f(2)>f(1)\) is false as no \(f(x)\) is more than \(f(1)\). (The correct would be \(f(2)<f(1)\)).
Hope it's clear.
HI Bunuel,
Please explain the red part , what i understood from making graph , if we move left from the axis point certainly value of f(x) decreases but if we move right side from the axis where x =1, value of f(x) will increase till certain point and then will decrease, so how can we deduce that value of f(x) is max at x =1
Answer to this question cannot be D, it should be B (II only).
To solve this question you don't need to calculate a, b, and c. We should put the given three points on the XY-plane and we'll get: parabola \(f(x)=ax^2+bx+c\) will have the vertex at \(x=1\), halfway between \(x=-3\) and \(x=5\), as \(f(-3)=0=f(5)\). Plus, as \(f(0)=3\) the parabola would be downward. As the parabola is downward, value of f(x) at x=1, f(1), is the the greatest value of f(x). As we move from x=1 to either of direction the value of f(x) will decrease. So any \(f(m)\) would be greater than \(f(n)\) if \(m\) is closer to \(x=1\) than \(n\). For example \(f(9)>f(10)>f(-12)\) or \(f(6)=f(-4)\) or \(f(-5)=f(7)>f(-100)\).
Hence I. \(f(-1)>f(2)\) is false, as \(x=2\) is 1 farther from \(x=1\) and \(x=-1\) is 2 farther than \(x=1\). (True inequality would be \(f(-1)<f(2)\)).
II. \(f(1)>f(0)\) is true as \(f(1)\) is vertex and more than any \(f(x)\).
III. \(f(2)>f(1)\) is false as no \(f(x)\) is more than \(f(1)\). (The correct would be \(f(2)<f(1)\)).
Hope it's clear.
Wonderful approach! thanks a lot!
gmatclubot
Re: If the parabola represented by f(x) = ax^2 + bx + c passes
[#permalink]
08 Oct 2015, 05:35
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