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Question Number 85262 by Rio Michael last updated on 20/Mar/20

$$\mathrm{find}\mathrm{the}\mathrm{centre}\mathrm{of}\mathrm{symmetry}\mathrm{of}\mathrm{the}\mathrm{curve}$$$$y=\frac{1}{x+2}$$

Commented by john santu last updated on 20/Mar/20

$$\mathrm{y}=-\mathrm{x}-2$$$$\mathrm{typo}$$

Commented by Rio Michael last updated on 20/Mar/20

$$howcomesir$$

Commented by mr W last updated on 20/Mar/20

$$symmetrycenteris(-2,0)$$$$symmetryaxesare$$$$y=x+2and$$$$y=-x-2$$

Commented by Rio Michael last updated on 20/Mar/20

$$\mathrm{is}\mathrm{the}\mathrm{centre}\mathrm{of}\mathrm{symmetry}\mathrm{thesame}\mathrm{as}$$$$\mathrm{the}\mathrm{vertical}\mathrm{asymptote}$$

Commented by mr W last updated on 20/Mar/20

$$noallsymmetricalcurveshavea$$$$asymptote,e.g.anellipseoraparabola.$$$$anellipsehastwosymmetryaxes,$$$$thereforealsoasymmetrycenter.$$$$aparabolahasonlyonesymmetry$$$$axis,thereforeithasnosymmetry$$$$center.$$

Commented by Rio Michael last updated on 20/Mar/20

$$\mathrm{thanks}\mathrm{alot}\mathrm{sir}$$

Answered by john santu last updated on 20/Mar/20

Commented by john santu last updated on 20/Mar/20

$$\mathrm{center}\mathrm{point}(-2,0)$$$$$$

Commented by mr W last updated on 20/Mar/20

Commented by Rio Michael last updated on 20/Mar/20

$$\mathrm{thanks}$$

Commented by mr W last updated on 20/Mar/20

$$ifacurvehastwoormoresymmetry$$$$axes,thentheintersectionpointof$$$$thesymmetryaxesisthesymmetry$$$$center.$$

Commented by jagoll last updated on 20/Mar/20

$$\mathrm{what}\mathrm{generally}\mathrm{formula}\mathrm{to}\mathrm{find}$$$$\mathrm{the}\mathrm{symetri}?$$

Commented by jagoll last updated on 20/Mar/20

$$\mathrm{if}\mathrm{y}=\frac{2}{{\mathrm{x}}^2-1},\mathrm{what}\mathrm{the}\mathrm{symetri}\mathrm{sir}$$

Commented by jagoll last updated on 20/Mar/20

$$\mathrm{the}\mathrm{graph}\mathrm{have}1\mathrm{axes}\mathrm{symetri}$$$$$$

Commented by mr W last updated on 20/Mar/20

$$therearesomeeasycasesto‘‘{see}^{″}the$$$$symmetry:$$$$curvey=f\left(x\right)$$$$iff(a+x)=f(a-x),thenx=aisa$$$$symmetryaxis.$$$$e.g.y=f\left(x\right)=\frac{2}{x^2-1},$$$$f\left(x\right)=f(-x)\Rightarrow x=0isasymmetryaxis.$$$$$$$$curvex=f\left(y\right)$$$$iff(b+y)=f(b-y),theny=bisa$$$$symmetryaxis.$$

Commented by jagoll last updated on 20/Mar/20

$$\mathrm{ok}\mathrm{sir}.\mathrm{thank}\mathrm{you}$$

Commented by mr W last updated on 20/Mar/20

$$generalcase:$$$$saywehaveacurveF(x,y)=0,$$$$ifF(x-\frac{2a(ax+by+c)}{a^2+b^2},y-\frac{2b(ax+by+c)}{a^2+b^2})=F(x,y),$$$$thenax+by+c=0isasymmetryaxis.$$$$why?$$$$seeQ84531$$$$ifF(x-\frac{2a(ax+by+c)}{a^2+b^2},y-\frac{2b(ax+by+c)}{a^2+b^2})=F(x,y),$$$$thatmeanstheimageofcurveF(x,y)$$$$inthelineax+by+c=0isthecurve$$$$itself,thisisthecasewhenthecurve$$$$issymmetricabouttheline$$$$ax+by+c=0.$$

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