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September 30, 2014

September 30, 2014

Posted by **ladybug** on Friday, October 26, 2012 at 9:29pm.

x^2+y^2+4x+2y-20=0

x^2+4x+4+y^2-2y+1=20+4+1

(x+2)^2+(y+1)^2=25--this is where you helped me. Then I came and did the x and y intercepts. This what I got. But not a coordinate???

x-intercepts are found by setting y=0

(x+2)^2+(0-1)^2=5^2

x^2+4x+4+1=25

x^2+4x-20=0

x=[-4+/-sqrt((4^2)-(4)(1)(-20))]/2

=[-4+/-sqrt(96)]/2

[-2+/-2sqrt(6)]

y-intercepts are found by setting x=0

(0+2)^2 +(y-1)^2=5^2

y^2-2y+1+4=25

y^2-2y-20=0

y=[2+/-sqrt((-2^2)-(4)(1)(-20))]/2

=[1+/-sqrt(21)]

- college algebra--Please help Reiny please -
**Reiny**, Friday, October 26, 2012 at 11:51pmWhat you did is correct, but it would have been easier to do it this way:

let y = 0

(x+2)^2 + 1 = 25

(x+2)^2 = 24

x + 2 = ± √24 = ± 2√6

x = -2 ± 2√6

same way for the y-intercept

so the x-intercepts are:

(-2+2√6 , 0) and (-2 - 2√6, 0)

the y-intercepts are:

(0, 1+√21) and (0,1 - √21)

Your prof seem "picky"

By stating it as the y-intercept, you are implying that x = 0

etc

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