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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