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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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02 Mar 2010, 08:10
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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, 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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02 Mar 2010, 10:54
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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]

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02 Mar 2010, 18:32
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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]

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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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03 Mar 2010, 00:18
Expert's post
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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]

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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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23 Dec 2013, 21:20
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Re: If the parabola represented by f(x) = ax^2 + bx + c passes [#permalink]

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25 Mar 2015, 10:07
Hello from the GMAT Club BumpBot!

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

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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
[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]

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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
[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 [ 11.58 KiB | Viewed 2955 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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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
[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.

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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10 Nov 2016, 08:47
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If the parabola represented by f(x) = ax^2 + bx + c passes   [#permalink] 10 Nov 2016, 08:47
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