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apoorvasrivastva
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If the parabola represented by f(x) = ax^2 + bx + c passes Updated on: 26 May 2015, 05:36
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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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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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
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B
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If the parabola represented by f(x) = ax^2 + bx + c passes 02 Mar 2010, 18:32
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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

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

OA is
Show Spoiler
D
Show more

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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If the parabola represented by f(x) = ax^2 + bx + c passes 02 Mar 2010, 10:54
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apoorvasrivastva


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)
Show more

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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If the parabola represented by f(x) = ax^2 + bx + c passes 03 Mar 2010, 00:01
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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 :(
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If the parabola represented by f(x) = ax^2 + bx + c passes 03 Mar 2010, 00:18
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apoorvasrivastva
@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 :(
Show more

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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If the parabola represented by f(x) = ax^2 + bx + c passes 04 Mar 2010, 21:22
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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.
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If the parabola represented by f(x) = ax^2 + bx + c passes 25 May 2015, 18:22
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Bunuel
apoorvasrivastva
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
Show 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.
Show more

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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If the parabola represented by f(x) = ax^2 + bx + c passes 26 May 2015, 05:51
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Bunuel
apoorvasrivastva
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
Show 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
Show more

Check below:
Attachment:
parabola.png
parabola.png [ 11.58 KiB | Viewed 11539 times ]

Also check parabola chapter HERE.

Hope it helps.
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If the parabola represented by f(x) = ax^2 + bx + c passes 08 Oct 2015, 06:35
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Bunuel
apoorvasrivastva
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
Show 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.
Show more


Wonderful approach! thanks a lot! :)
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If the parabola represented by f(x) = ax^2 + bx + c passes 07 Jan 2018, 09:46
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apoorvasrivastva
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
Show 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.
Show more


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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Bunuel
apoorvasrivastva
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
Show 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, 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.
Show more

Hint: put the given three points on the plane,
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If the parabola represented by f(x) = ax^2 + bx + c passes 07 Jan 2018, 10:25
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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 the parabola represented by f(x) = ax^2 + bx + c passes 23 Sep 2020, 23:10
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Concept: Once you find the Line of Symmetry of a Downward Opening Parabola, the X Coordinate at this Vertex will give you the HIGHEST VALUE that f(x) = y can have


Rule: to find the Line of Symmetry of a Parabola (in which everything on the Left Side of the Axis/Line is a "MIRROR REFLECTION" of all the Points on the Right Side), find the MID-POINT between the X-Intercepts that cross the X-Axis


between (-3 , 0) and (+5 , 0) ---- the Line of Symmetry will occur at the Vertical Line of: X = +1

Also, since this is a Downward Opening Parabola, the Vertex at f(+1) will also provide the HIGHEST VALUE of y

I. f(-1) > f(2)


f(-1) is +2 Units to the LEFT of the Highest Value at the Vertex f(+1)

However, f(2) is only +1 Unit to the RIGHT of the Highest Value at the Vertex f(+1)

the Value of f(2) will be GREATER than the Value of f(-1)

NOT True



II. f(1) > f(0)

The Highest Value will occur at the Vertex, when X =+1, because this is a Downward Opening Parabola.

Thus f(1) will be greater than any other Value

TRUE


III. f(2) > f(1)

As stated above, the Highest Value will occur at the Vertex when the X-Coordinate is +1 (at the Line of Symmetry) because this is a Downward Opening Parabola.

thus: f(2) < f(1)

III is NOT True


ONLY II must be True

-B-
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