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Manager  Joined: 29 Jul 2009
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If the parabola represented by f(x) = ax^2 + bx + c passes  [#permalink]

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Question Stats: 48% (02:54) correct 52% (02:39) wrong based on 472 sessions

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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.
Math Expert V
Joined: 02 Sep 2009
Posts: 59628
Re: If the parabola represented by f(x) = ax^2 + bx + c passes  [#permalink]

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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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Intern  Joined: 22 Nov 2009
Posts: 16
Re: If the parabola represented by f(x) = ax^2 + bx + c passes  [#permalink]

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2
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!)
##### General Discussion
Manager  Joined: 29 Jul 2009
Posts: 73
Re: If the parabola represented by f(x) = ax^2 + bx + c passes  [#permalink]

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@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 Math Expert V
Joined: 02 Sep 2009
Posts: 59628
Re: If the parabola represented by f(x) = ax^2 + bx + c passes  [#permalink]

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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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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.
Manager  Joined: 03 May 2013
Posts: 68
Re: If the parabola represented by f(x) = ax^2 + bx + c passes  [#permalink]

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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
Math Expert V
Joined: 02 Sep 2009
Posts: 59628
Re: If the parabola represented by f(x) = ax^2 + bx + c passes  [#permalink]

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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 [ 11.58 KiB | Viewed 5229 times ]

Also check parabola chapter HERE.

Hope it helps.
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Manager  Joined: 02 Jul 2015
Posts: 101
Schools: ISB '18
GMAT 1: 680 Q49 V33 Re: If the parabola represented by f(x) = ax^2 + bx + c passes  [#permalink]

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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! Manager  B
Status: Turning my handicaps into assets
Joined: 09 Apr 2017
Posts: 121
Re: If the parabola represented by f(x) = ax^2 + bx + c passes  [#permalink]

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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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If time was on my side, I'd still have none to waste......
Math Expert V
Joined: 02 Sep 2009
Posts: 59628
Re: If the parabola represented by f(x) = ax^2 + bx + c passes  [#permalink]

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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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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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If time was on my side, I'd still have none to waste...... Re: If the parabola represented by f(x) = ax^2 + bx + c passes   [#permalink] 07 Jan 2018, 10:25
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