How do you determine whether the graph of #x+y^2=1# is symmetric with respect to the x axis, y axis, the line y=x or y=-x, or none of these?

2 Answers
Nov 26, 2017

Answer:

#x+y^2=1# is symmetric wrt the #x-#axis

Explanation:

Let's look at the graph of #x+y^2=1# in relation to the #x-# and #y-#axes and the lines #y=x and y=-x# below.

graph{(x+y^2-1)(y-x)(y+x)=0 [-10, 10, -5, 5]}

Clearly, #x+y^2=1# is symmetric wrt the #x-#axis bur not the other options presented in this question.

Nov 26, 2017

Answer:

The curve is only symmetric w.r.t. #x#-axis.

Explanation:

If a graph is symmetric w.r.t. #x# axis, then for every #(x_a,y_a)# on the curve, #(x_a,-y_a)# too lies on the curve.

As #(x_a,y_a)# lies on curve #x+y^2=1#, we have #x_a+y_a^2=1#

an hence #x_a+(-y_a)^2=x_a^2+y_a^2=1# and #(x_a,-y_a)# too lies on curve and hence #x+y^2=1# is symmetric w.r.t. #x#-axis.

For being symmetric w.r.t. #y#-axis, we should have #(-x_a,y_a)# too on curve. Here #-x_a+y_a^2!=1#, hence curve is not symmetric w.r.t. #y#-axis.

For being symmetric w.r.t. #y=x#, #(y_a,x_a)# too should be on the curve, but again #y_a+x_a^2!=1#, hence curve is not symmetric w.r.t. #y=x#.

For being symmetric w.r.t. #y=-x#, #(-y_a,-x_a)# too should be on the curve, but again #-y_a+(-x_a)^2=x_a^2-y_a!=1#, hence curve is not symmetric w.r.t. #y=-x#.

graph{x+y^2=1 [-14.71, 5.29, -4.8, 5.2]}