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

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On the x-y coordinate plane, a point (1,2) is on a parabola y=ax^2+bx+  [#permalink]

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New post 31 Jan 2016, 21:39
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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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On the x-y coordinate plane, a point (1,2) is on a parabola y=ax^2+bx+  [#permalink]

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New post 31 Jan 2016, 23:31
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
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Re: On the x-y coordinate plane, a point (1,2) is on a parabola y=ax^2+bx+  [#permalink]

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New post 02 Feb 2016, 17:25
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.
Therefore, the answer is D.
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Re: On the x-y coordinate plane, a point (1,2) is on a parabola y=ax^2+bx+  [#permalink]

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New post 02 Jan 2018, 21:21
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

My answer is D

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

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New post 02 Jan 2018, 22:13
y=ax2 + bx +c = f(x) (a not equal to 0)

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

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

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

Each stmt in itself satisfies
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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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