# How do you determine whether the graph of 6x^2=y-1 is symmetric with respect to the x, y axis, the line y=x, the line y=-x, or none of these?

May 10, 2017

It is symmetric only w.r.t. $y$-axis.

#### Explanation:

1 $-$ If a graph represented by $f \left(x , y\right) = 0$ is symmetric with respect to $x$-axis, we should have $f \left(x , y\right) = f \left(x , - y\right)$.

Here, in $6 {x}^{2} = y - 1$, we have $f \left(x , y\right) = 6 {x}^{2} - y + 1 = 0$ and $f \left(x , - y\right) = 6 {x}^{2} - \left(- y\right) + 1 = 6 {x}^{2} + y + 1$ and hence

$f \left(x , y\right) \ne f \left(x , - y\right)$ and hence it is **not symmetric w.r.t. $x$-axis.

2 $-$ If a graph represented by $f \left(x , y\right) = 0$ is symmetric with respect to $y$-axis, we should have $f \left(x , y\right) = f \left(- x , y\right)$.

In $6 {x}^{2} = y - 1$, we have $f \left(- x , y\right) = 6 {\left(- x\right)}^{2} - y + 1 = 6 {x}^{2} - y + 1$ and hence

$f \left(x , y\right) = f \left(- x , y\right)$ and hence it is symmetric w.r.t. $y$-axis.

3 $-$ If a graph represented by $f \left(x , y\right) = 0$ is symmetric with respect to line $y = x$, we should have $f \left(x , y\right) = f \left(y , x\right)$.

In $6 {x}^{2} = y - 1$, we have $f \left(y , x\right) = 6 {y}^{2} - x + 1$ and hence

$f \left(x , y\right) \ne f \left(y , x\right)$ and hence it is not symmetric w.r.t. line $y = x$.

4 $-$ If a graph represented by $f \left(x , y\right) = 0$ is symmetric with respect to line $y = - x$, we should have $f \left(x , y\right) = f \left(- y , - x\right)$.

In $6 {x}^{2} = y - 1$, we have $f \left(- y , - x\right) = 6 {\left(- y\right)}^{2} - \left(- x\right) + 1 = 6 {y}^{2} + x + 1$ and hence

$f \left(x , y\right) \ne f \left(- y , - x\right)$ and hence it is not symmetric w.r.t. line $y = - x$.

graph{(6x^2-y+1)(x-y)(x+y)=0 [-5.394, 4.606, -0.64, 4.36]}