Graph x=(y+2)^2

Problem

x=(y+2)2

Solution

1. Identify the type of conic section. Since the equation is in the form x=a*(y−k)2+h it represents a parabola that opens horizontally to the right.

2. Determine the vertex (h,k) By comparing x=1*(y−(−2))2+0 to the standard form, the vertex is (0,−2)

3. Find the xintercept by setting y=0

x=(0+2)2

x=4

The xintercept is (4,0)
4. Find the yintercepts by setting x=0

0=(y+2)2

0=y+2

y=−2

The yintercept is (0,−2) which is also the vertex.
5. Select additional points to determine the width. If y=−1

x=(−1+2)2

x=1

If y=−3

x=(−3+2)2

x=1

The points (1,−1) and (1,−3) are on the graph.
6. Sketch the curve through the points (0,−2) (1,−1) (1,−3) and (4,0) ensuring it is a smooth curve opening to the right with the axis of symmetry y=−2

Final Answer

x=(y+2)2* is a right-opening parabola with vertex *(0,−2)

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