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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]

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New post Updated on: 26 May 2015, 05:36
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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

Originally posted by apoorvasrivastva on 02 Mar 2010, 08:10.
Last edited by Bunuel on 26 May 2015, 05:36, 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]

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New post 02 Mar 2010, 18:32
10
14
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


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]

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New post 02 Mar 2010, 10:54
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3
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]

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New post 03 Mar 2010, 00:01
@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]

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New post 03 Mar 2010, 00:18
1
3
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]

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New post 04 Mar 2010, 21:22
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]

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New post 25 May 2015, 18:22
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


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]

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New post 26 May 2015, 05:51
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


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 5229 times ]


Also check parabola chapter HERE.

Hope it helps.
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Re: If the parabola represented by f(x) = ax^2 + bx + c passes  [#permalink]

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New post 08 Oct 2015, 06:35
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


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! :)
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Re: If the parabola represented by f(x) = ax^2 + bx + c passes  [#permalink]

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New post 07 Jan 2018, 09:46
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


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, could you please explain the highlighted part. How do we infer that parabola is downward? According to theory we only know that parabola is downward if a<0.
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Re: If the parabola represented by f(x) = ax^2 + bx + c passes  [#permalink]

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New post 07 Jan 2018, 09:49
Mehemmed 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


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, could you please explain the highlighted part. How do we infer that parabola is downward? According to theory we only know that parabola is downward if a<0.


Hint: put the given three points on the plane,
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Re: If the parabola represented by f(x) = ax^2 + bx + c passes  [#permalink]

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New post 07 Jan 2018, 10:25
I got it. We can know it when putting all the 3 points on a plane. I thought there's another special way. Thanks
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Re: If the parabola represented by f(x) = ax^2 + bx + c passes   [#permalink] 07 Jan 2018, 10:25
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