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In Problems 65-70, graph each equation and find the point(s) of intersection, if any.

The circlex-12+y+22=4and the parabolay2+4y-x+1=0.

Short Answer

Expert verified

The graph of the parabola y2+4y-x+1=0and the circle x-12+y+22=4is,

From the graph, the two equations are intersecting at two points They are,0,-3-2,1,0,0,3-2,1,-4.

Step by step solution

01

Step 1. Given information

y2+4y-x+1=0x-12+y+22=4

First solve each equation for y.

Consider the first equation,

y2+4y-x+1=0y2+4y=x-1y2+4y+4=x-1+4y+22=x+3y+2=±x+3y=±x+3-2

Next consider the second equation,

x-12+y+22=4y+22=4-x-12y+2=±4-x-12y=±4-x-12-2

02

Step 2. Now sketch a graph by using the obtained equations.

From the graph, the two equations are intersecting at two points. They are,0,-3-2,1,0,0,3-2,1,-4.

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