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sketch y=x^2 and then on separate axis sketch...a) y=2x^2 b) y=(x+1)^2 -3 c)...
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High School Teacher
y = x^2.
This a parabola.
y = 0 when x = 0. So (x,y) = (0, 0) vertex.
x = 0 is axis of symmetry of the paraboa.
y= x^2 could be written as x^2 = 4ay Or x^2 = 4(1/4)y. So 1/4 is focal length of the parabola . The focus is at (0,1/4).
Since x^2 is always positive, y is also positive. So the curve is above X axis . Both branches approach positive infinity as x --> infinity or minus infinity.
2)y = 2x^2
The parabola has vetex (0,0).
x^2 = y/2. Or x^2 = 4(1/8)x. So 1/8 is the focal length.
(0,1/8) is the coordinate of the focus.
The parabola is open upward and above X axis.
y = (x+1)^2-3 Or
y-(-3) = (x-(-1))^2 is a parabola with vertex at ( -3, -1).
The vetex i below x axis.
x+1 = 0 , Or x =-1 is th axis of symmetry of the parabola.
The parobola intercepts y axis at y = (0-(-1)^3 -3 = -2.
The parabola intercepts xaxis at -1+sqt3 and at -1-sqrt3.
The parabla is open upward.
c) y = -(x-2)^2 +1.
Or y-1 = -(x-2)^2 is a parabola with vertex at (2,1).
x-2 = 0 Or x= 2 is the axis of symmetry of the parabola.
The parabola intercepts y axis at y = -(0-2)^2+1 = -3.
The parabola intersects x axis at the zeros of -(x-2)^2+1:
So ( x-2)^2 = 1. Or x -2 = 1 Or x -2 = -1.
x= 2+1 =3 Or x = 2-1 = 1 are the two points intercepts of x axis.
For y > 0 for any x inthe interval (1 , 3)
For all x> 3 and for all x< 1, y is negative.
So the parabola is open down ward going infinity as xapprache + or -infinity.
Posted by neela on October 28, 2010 at 8:08 PM (Answer #1)
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