Question

Determine algebraically whether the graph of equations is symmetric to the x-axis, y-axis, origin or none of these.
A). $y=|x|+5$
B). $xy=4$
C). $x+{{y}^{2}}=8$

Answer 1

Type of question is based on the concept of finding symmetry algebraically, which says that If we want to find any graph symmetric to x-axis, then replace ‘y’ with ‘-y’ and on simplifying if we get same equation we have in initial then it is symmetric, otherwise not. Same it with for symmetric about y-axis, replace ‘x’ with ‘-x’ if we get the equation we have initially then it is symmetric otherwise not. And to check for symmetry about origin replace both ‘x’ and ‘y’ with ‘-x’ and ‘-y’ respectively and on simplifying we get the equation we have initially then it is symmetric about origin otherwise not.

Complete step-by-step solution:
Moving ahead with the question, first let us check symmetricity for $y=|x|+5$;
Check for symmetric about y-axis;
Replace ‘x’ with ‘-x’ then we will get;
$\begin{aligned} & y=|x|+5 \\\ & y=|-x|+5 \\\ \end{aligned}$
As we know that $|-a|=|a|$, in which ‘||’ represents mode.
So we will get;
$y=|x|+5$
Hence the graph is symmetric about the y-axis.
Check for symmetric about x-axis;
Replace ‘y’ with ‘-y’ then we will get;
$\begin{aligned} & y=|x|+5 \\\ & -y=|x|+5 \\\ \end{aligned}$
So we will get;
$-y=|x|+5$
As this equation is not same as $y=|x|+5$, so we can say that the graph is not symmetric about the x-axis.
Similarly check for symmetric about origin;
Replace both ‘x’ and ‘y’ with ‘-x’ and ‘-y’ respectively
So we will get;
$\begin{aligned} & y=|x|+5 \\\ & -y=|-x|+5 \\\ \end{aligned}$
Which we can reduce it to;
$-y=|x|+5$
As this equation is not same as $y=|x|+5$, so we can say that the graph is not symmetric about origin.
B). Now let us check symmetricity about $xy=4$;
Check for symmetric about y-axis;
Replace ‘x’ with ‘-x’ then we will get;
$\begin{aligned} & xy=4 \\\ & \left( -x \right)y=4 \\\ \end{aligned}$
So we can reduce it to;
$-xy=4$
As this equation is not the same as $xy=4$, hence the graph is not symmetric about y-axis.
Check for symmetric about x-axis;
Replace ‘y’ with ‘-y’ then we will get;
$\begin{aligned} & xy=4 \\\ & x\left( -y \right)=4 \\\ \end{aligned}$
So we can reduce it to;
$-xy=4$
As this equation is not the same as $xy=4$, so we can say that the graph is not symmetric about the x-axis.
Similarly check for symmetric about origin;
Replace both ‘x’ and ‘y’ with ‘-x’ and ‘-y’ respectively
So we will get;
$\begin{aligned} & xy=4 \\\ & \left( -x \right)\left( -y \right)=4 \\\ \end{aligned}$
SO we can reduce it to;
$xy=4$
As this equation is the same as $xy=4$, so we can say that the graph is symmetric about origin.
C). Now let us check symmetricity for $x+{{y}^{2}}=8$;
Check for symmetric about y-axis;
Replace ‘x’ with ‘-x’ then we will get;
$\begin{aligned} & x+{{y}^{2}}=8 \\\ & \left( -x \right)+{{y}^{2}}=8 \\\ \end{aligned}$
So we can reduce it to;
$-x+{{y}^{2}}=8$
As the equation is not the same as $x+{{y}^{2}}=8$ which is initially one, Hence the graph is not symmetric about the y-axis.
Check for symmetric about x-axis;
Replace ‘y’ with ‘-y’ then we will get;
$\begin{aligned} & x+{{y}^{2}}=8 \\\ & x+{{\left( -y \right)}^{2}}=8 \\\ \end{aligned}$
So we will get;
$x+{{y}^{2}}=8$
As this equation is the same as $x+{{y}^{2}}=8$, so we can say that the graph is symmetric about the x-axis.
Similarly check for symmetric about origin;
Replace both ‘x’ and ‘y’ with ‘-x’ and ‘-y’ respectively
So we will get;
$\begin{aligned} & x+{{y}^{2}}=8 \\\ & \left( -x \right)+{{\left( -y \right)}^{2}}=8 \\\ \end{aligned}$
Which we can reduce it to;
$-x+{{y}^{2}}=8$
As this equation is not the same as $x+{{y}^{2}}=8$, so we can say that the graph is not symmetric about origin.
Hence from above result we can say that the equation $y=|x|+5$ is symmetric about x-axis, the equation $xy=4$ is symmetric about origin and the equation $x+{{y}^{2}}=8$ is symmetric about x-axis.

Note: This is the algebraic process of finding out the equation whether it is symmetric about origin, x-axis or y-axis. Otherwise we can find the symmetry by drawing the equation on the cartesian plane.