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If the parabola represented by f(x) = ax^2 + bx + c passes

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If the parabola represented by f(x) = ax^2 + bx + c passes [#permalink] New post 02 Mar 2010, 07:10
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If the parabola represented by f(x) = ax^2 + bx + c passes through points (-3,0) , (0,3) and (5,0), which of the following must be true?

I. f(-1) > f(2)
II. f(1) > f(0)
III. f(2) > f(1)

A. Only I
B. Only II
C. Only III
D. I and II
E. I and III
[Reveal] Spoiler: OA

Last edited by Bunuel on 26 May 2015, 04:36, edited 4 times in total.
Edited the OA.
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Re: If the parabola represented by f(x) = ax^2 + bx + c passes [#permalink] New post 02 Mar 2010, 09:54
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apoorvasrivastva wrote:

If the parabola represented by f(x) = ax^2 + bx + c passes through points (-3,0) , (0,3) and (5,0), which of the following must be true?

I f(-1) > f(2)
II f(1) > f(0)
III f(2) > f(1)


f(x)=y

Substituting (x,y) as (-3,0) , (0,3) and (5,0), we get the following equations:

0 = 9a-3b+c
3 = c
0 = 25a+5b+c

Solving: a=-1/5, b=2/5, c=3
f(x) = -x^2/5 + x/5 +3

The options are on f(-1), f(0), f(1) and f(2).
f(-1) = -1/5 + 2/5 + 3 = 16/5 = 3.2
f(0) = 3
f(1) = 1/5 + 2/5 + 3 = 18/5 = 3.6
f(2) = -4/5 + 4/5 + 3 = 3

I f(-1) > f(2) true
II f(1) > f(0) true
III f(2) > f(1) false

So, option D, I and II

(not sure if there is an easier way to solve this!)
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Re: If the parabola represented by f(x) = ax^2 + bx + c passes [#permalink] New post 02 Mar 2010, 17:32
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apoorvasrivastva wrote:
If the parabola represented by f(x) = ax^2 + bx + c passes through points (-3,0) , (0,3) and (5,0), which of the following must be true?

I f(-1) > f(2)
II f(1) > f(0)
III f(2) > f(1)

A. only I
B. only II
c. only III
D. I and II
E. I and III

please suggest me as to how do i crack this question in 2 mints..i am completely lost on this :(

OA is
[Reveal] Spoiler:
D


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.
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Re: If the parabola represented by f(x) = ax^2 + bx + c passes [#permalink] New post 02 Mar 2010, 23:01
@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 :(
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Re: If the parabola represented by f(x) = ax^2 + bx + c passes [#permalink] New post 02 Mar 2010, 23:18
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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 \(f(-1)>f(-2)\).
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Re: If the parabola represented by f(x) = ax^2 + bx + c passes [#permalink] New post 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.
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Re: If the parabola represented by f(x) = ax^2 + bx + c passes [#permalink] New post 23 Dec 2013, 20:20
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Re: If the parabola represented by f(x) = ax^2 + bx + c passes [#permalink] New post 25 Mar 2015, 09:07
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Re: If the parabola represented by f(x) = ax^2 + bx + c passes [#permalink] New post 25 May 2015, 17:22
Bunuel wrote:
apoorvasrivastva wrote:
If the parabola represented by f(x) = ax^2 + bx + c passes through points (-3,0) , (0,3) and (5,0), which of the following must be true?

I f(-1) > f(2)
II f(1) > f(0)
III f(2) > f(1)

A. only I
B. only II
c. only III
D. I and II
E. I and III

please suggest me as to how do i crack this question in 2 mints..i am completely lost on this :(

OA is
[Reveal] Spoiler:
D


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
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Re: If the parabola represented by f(x) = ax^2 + bx + c passes [#permalink] New post 26 May 2015, 04:51
Expert's post
vipulgoel wrote:
Bunuel wrote:
apoorvasrivastva wrote:
If the parabola represented by f(x) = ax^2 + bx + c passes through points (-3,0) , (0,3) and (5,0), which of the following must be true?

I f(-1) > f(2)
II f(1) > f(0)
III f(2) > f(1)

A. only I
B. only II
c. only III
D. I and II
E. I and III

please suggest me as to how do i crack this question in 2 mints..i am completely lost on this :(

OA is
[Reveal] Spoiler:
D


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


Check below:
Attachment:
parabola.png
parabola.png [ 11.58 KiB | Viewed 679 times ]


Also check parabola chapter HERE.

Hope it helps.
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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis ; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) ; 12. Tricky questions from previous years.

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: If the parabola represented by f(x) = ax^2 + bx + c passes   [#permalink] 26 May 2015, 04:51
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