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May 5, 2016

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

You helped me with this question and this is what I got but my Professor says that each x and y intercept needs to have a coordinate. Where did I go wrong?? Can you help me?

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 11:51pm

What 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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