Question

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?

Answer 1

`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.

Answer 2

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]}