SOLUTION: 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

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 Click here to see ALL problems on Quadratic-relations-and-conic-sections 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(6742)   (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 ** (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.