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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 7612
GMAT 1: 760 Q51 V42 GPA: 3.82
On the x-y coordinate plane, a point (1,2) is on a parabola y=ax^2+bx+  [#permalink]

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Difficulty:   75% (hard)

Question Stats: 50% (01:48) correct 50% (01:45) wrong based on 116 sessions

### HideShow timer Statistics On the x-y coordinate plane, a point (1,2) is on a parabola y=ax^2+bx+c (a≠0), is a point (-1,2) on the parabola, too?

1) b=0
2) f(x)=f(-x)

* A solution will be posted in two days.

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Math Expert V
Joined: 02 Aug 2009
Posts: 7763
On the x-y coordinate plane, a point (1,2) is on a parabola y=ax^2+bx+  [#permalink]

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MathRevolution wrote:
On the x-y coordinate plane, a point (1,2) is on a parabola y=ax^2+bx+c (a≠0), is a point (-1,2) on the parabola, too?

1) b=0
2) f(x)=f(-x)

* A solution will be posted in two days.

Hi,

here we have a parabola...
some properties of parabola that will help us nail the Q..
1) the max/min of the parabola occurs at -b/2a..
2) the point -b/2a acts as axis of symmetry...
3) If a in ax^2 is positive, the parabola opens upwards
4) If a in ax^2 is negative, the parabola opens downwards
...

lets see the Q stem for info now..
point (1,2) is on the parabola y=ax^2+bx+c (a≠0)..
parabola can open upwards or downwards as a can be positive/-ive..
IS (-1,2) also on the parabola...
so one way is to see if they both lie on either side of axis of symmetry..

lets see the statements..

1) b=0
since b=0, -b/2a=0 or the parabola has its vertex at x=0...
when x=0, y=c..
so axis of symmetry is y-axis..
so (-1,2) will be on parabola as (1,2) is on parabola
suff

2) f(x)=f(-x)
f(x)=ax^2+bx+c..
f(-x)=a(-x)^2+b(-x)+c=ax^2-bx+c..

if f(x)=f(-x), it is an even function and even function is symmetric about y axis..
same as above..
suff

D

SIDE NOTE:- even /odd functions are generally not seen too often
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 7612
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: On the x-y coordinate plane, a point (1,2) is on a parabola y=ax^2+bx+  [#permalink]

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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

On the x-y coordinate plane, a point (1,2) is on a parabola y=ax^2+bx+c (a≠0), is a point (-1,2) on the parabola, too?

1) b=0
2) f(x)=f(-x)

When you modify the original condition and the question, the question is if y=ax^2+bx+c is symmetry of the y-axis as (1,2) passes parabola and the question asks if (-1,2) passes parabola as well.
1)=2) the both means y=ax^2+bx+c is symmetry of the y-axis, which is yes.
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Manager  B
Joined: 17 Sep 2017
Posts: 54
Re: On the x-y coordinate plane, a point (1,2) is on a parabola y=ax^2+bx+  [#permalink]

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2
Quote:
On the x-y coordinate plane, a point (1,2) is on a parabola y=ax^2+bx+c (a≠0), is a point (-1,2) on the parabola, too?

1) b=0
2) f(x)=f(-x)

Point (1,2) is on f(x) => 2 = a + b + c .
Is point (-1,2) on f(x) => 2 = a - b + c or else a + b + c = a - b + c ? or else b = 0?

A. b=0 => Suf
B. f(x) = f(-x)
=> ax^2 + bx + c = ax^2 - bx+c => 2bx = 0 => b=0 => Suf

Experts, could you please clarify whether I'm doing it right? Many thanks Manager  B
Joined: 23 Oct 2017
Posts: 61
Re: On the x-y coordinate plane, a point (1,2) is on a parabola y=ax^2+bx+  [#permalink]

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y=ax2 + bx +c = f(x) (a not equal to 0)

since point (1,2) lies on y=f(x)
2= a+b+c - 

Now does (-1,2) lie on y= f(x) ? i.e. is 2= a-b+c ? -
(on substituing -1,2 in the parabola equation)

stmt 1: b=0 eqn  is satsified
stmt 2: f(x)=f(-x) => a+b+c = a-b +c
=> b=0
Thus eqn  is satsified

Each stmt in itself satisfies
Non-Human User Joined: 09 Sep 2013
Posts: 11720
Re: On the x-y coordinate plane, a point (1,2) is on a parabola y=ax^2+bx+  [#permalink]

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_________________ Re: On the x-y coordinate plane, a point (1,2) is on a parabola y=ax^2+bx+   [#permalink] 26 Mar 2019, 07:13
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