# sketch y=x^2 and then on separate axis sketch...a) y=2x^2 b) y=(x+1)^2 -3 c) y=-(x-2)^2+1

neela | Student

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.

3)

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.