# 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?

Nov 26, 2017

$x + {y}^{2} = 1$ 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 $y = x \mathmr{and} y = - x$ 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

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

#### Explanation:

If a graph is symmetric w.r.t. $x$ axis, then for every $\left({x}_{a} , {y}_{a}\right)$ on the curve, $\left({x}_{a} , - {y}_{a}\right)$ too lies on the curve.

As $\left({x}_{a} , {y}_{a}\right)$ lies on curve $x + {y}^{2} = 1$, we have ${x}_{a} + {y}_{a}^{2} = 1$

an hence ${x}_{a} + {\left(- {y}_{a}\right)}^{2} = {x}_{a}^{2} + {y}_{a}^{2} = 1$ and $\left({x}_{a} , - {y}_{a}\right)$ 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 $\left(- {x}_{a} , {y}_{a}\right)$ too on curve. Here $- {x}_{a} + {y}_{a}^{2} \ne 1$, hence curve is not symmetric w.r.t. $y$-axis.

For being symmetric w.r.t. $y = x$, $\left({y}_{a} , {x}_{a}\right)$ too should be on the curve, but again ${y}_{a} + {x}_{a}^{2} \ne 1$, hence curve is not symmetric w.r.t. $y = x$.

For being symmetric w.r.t. $y = - x$, $\left(- {y}_{a} , - {x}_{a}\right)$ too should be on the curve, but again $- {y}_{a} + {\left(- {x}_{a}\right)}^{2} = {x}_{a}^{2} - {y}_{a} \ne 1$, hence curve is not symmetric w.r.t. $y = - x$.

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