# 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

Verified

279.9k+ views

**Hint:**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{align}

& y=|x|+5 \\

& y=|-x|+5 \\

\end{align}$

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{align}

& y=|x|+5 \\

& -y=|x|+5 \\

\end{align}$

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{align}

& y=|x|+5 \\

& -y=|-x|+5 \\

\end{align}$

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{align}

& xy=4 \\

& \left( -x \right)y=4 \\

\end{align}$

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{align}

& xy=4 \\

& x\left( -y \right)=4 \\

\end{align}$

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{align}

& xy=4 \\

& \left( -x \right)\left( -y \right)=4 \\

\end{align}$

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{align}

& x+{{y}^{2}}=8 \\

& \left( -x \right)+{{y}^{2}}=8 \\

\end{align}$

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{align}

& x+{{y}^{2}}=8 \\

& x+{{\left( -y \right)}^{2}}=8 \\

\end{align}$

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{align}

& x+{{y}^{2}}=8 \\

& \left( -x \right)+{{\left( -y \right)}^{2}}=8 \\

\end{align}$

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.

Recently Updated Pages

Which of the following would not be a valid reason class 11 biology CBSE

What is meant by monosporic development of female class 11 biology CBSE

Draw labelled diagram of the following i Gram seed class 11 biology CBSE

Explain with the suitable examples the different types class 11 biology CBSE

How is pinnately compound leaf different from palmately class 11 biology CBSE

Match the following Column I Column I A Chlamydomonas class 11 biology CBSE

Trending doubts

What is 1 divided by 0 class 8 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

What is the past tense of read class 10 english CBSE

What is pollution? How many types of pollution? Define it

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

How many crores make 10 million class 7 maths CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

How fast is 60 miles per hour in kilometres per ho class 10 maths CBSE