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?

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Answer 1 · 2017-05-10T12:38:38

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

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

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

$f(x,y)!=f(x,-y)$ and hence it is **not symmetric w.r.t. $x$ -axis.

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

In $6x^2=y-1$ , we have $f(-x,y)=6(-x)^2-y+1=6x^2-y+1$ and hence

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

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

In $6x^2=y-1$ , we have $f(y,x)=6y^2-x+1$ and hence

$f(x,y)!=f(y,x)$ and hence it is not symmetric w.r.t. line $y=x$ .

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

In $6x^2=y-1$ , we have $f(-y,-x)=6(-y)^2-(-x)+1=6y^2+x+1$ and hence

$f(x,y)!=f(-y,-x)$ 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]}

How do you determine whether the graph of 6x^2=y-16x^2=y-1 is symmetric with respect to the x, y axis, the line y=x, the line y=-x, or none of these?