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Question 597524: Need help solving this -
1)(x-2)+y^2=16
It is wanting me to find the center, x- intercept y-intercept , solve for y and then name the curve of the graph
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! 1)(x-2)+y^2=16
It is wanting me to find the center, x- intercept y-intercept , solve for y and then name the curve of the graph
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(x-2)+y^2=16
x+y^2=18
y^2=-x+18
y^2=-(x-18)
This is an equation for a parabola that opens leftwards.
Its standard form: (y-k)^2=-4p(x-k), (h,k)=(x,y) coordinates of the vertex
x-intercept
set y=0
-x+18=0
x=18
..
y-intercept
set x=0
y^2=18
y=±√18=±3√2
..
There is no center for a parabola which this is.
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