Answer
The graph of the given function is shown below:
Work Step by Step
The provided function is,
$\frac{\left( x-5 \right)}{\left( 10x-2 \right)}\div \frac{{{x}^{2}}-10x+25}{25{{x}^{2}}-1}$
Simplify the above equation using the identity $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right)$:
$\begin{align}
& \frac{\left( x-5 \right)}{\left( 10x-2 \right)}\div \frac{{{x}^{2}}-10x+25}{25{{x}^{2}}-1}=\frac{\left( x-5 \right)}{2\left( 5x-1 \right)}\div \frac{{{x}^{2}}-5x-5x+25}{25{{x}^{2}}-1} \\
& =\frac{\left( x-5 \right)}{2\left( 5x-1 \right)}\div \frac{x\left( x-5 \right)-5\left( x-5 \right)}{{{\left( 5x \right)}^{2}}-{{1}^{2}}} \\
& =\frac{\left( x-5 \right)}{2\left( 5x-1 \right)}\div \frac{\left( x-5 \right)\left( x-5 \right)}{\left( 5x-1 \right)\left( 5x+1 \right)} \\
& =\frac{\left( 5x+1 \right)}{2\left( x-5 \right)}
\end{align}$
Then, the simplified expression is $\frac{\left( 5x+1 \right)}{2\left( x-5 \right)}$.
Hence, the equation for function $f$ is $f\left( x \right)=\frac{\left( 5x+1 \right)}{2\left( x-5 \right)}$.
To determine the symmetry, put $x$ equal to $-x$ , and find the symmetry of the function.
$\begin{align}
& f\left( -x \right)=\frac{\left( 5\left( -x \right)+1 \right)}{2\left( -x-5 \right)} \\
& =\frac{1-5x}{-2\left( x+5 \right)}
\end{align}$
The obtained value of $f\left( -x \right)$ is not equal to either function $f\left( x \right)$ or $-f\left( x \right)$. Therefore, the graph of the function is not symmetric with respect to the y-axis nor the origin.
To find the y-intercept, evaluate $f\left( 0 \right)$ such that
$\begin{align}
& f\left( 0 \right)=\frac{\left( \left( 5\times 0 \right)+1 \right)}{2\left( 0-5 \right)} \\
& =-\frac{1}{10}
\end{align}$
The y-intercept is $-\frac{1}{10}$ , so the graph passes through the point $\left( 0,-\frac{1}{10} \right)$.
To find the x-intercept, put the numerator equal to 0 as:
$\left( 5x+1 \right)=0$
Subtract $1$ from both sides of the equation,
$\begin{align}
& 5x+1-1=-1 \\
& 5x=-1
\end{align}$
Divide both sides of the equation by 5,
$\begin{align}
& \frac{5x}{5}=\frac{-1}{5} \\
& x=-\frac{1}{5}
\end{align}$
The x-intercept is $-\frac{1}{5}$ , so the graph passes through the point $\left( -\frac{1}{5},0 \right)$.
Now, find the vertical asymptote.
To find the vertical asymptote, put the denominator of the function equal to zero:
$2\left( x-5 \right)=0$
Divide both sides of the equation by 2:
$\begin{align}
& \frac{2\left( x-5 \right)}{2}=\frac{0}{2} \\
& x-5=0
\end{align}$
Add 5 to both sides of the equation:
$\begin{align}
& x-5+5=0+5 \\
& x=5
\end{align}$
Thus, the graph has a vertical asymptote and its equation is $x=5$.
Now, find the horizontal asymptote.
To find the horizontal asymptote, if the degrees of the denominator and numerator are the same, then divide the leading coefficient of the numerator by the leading coefficient of the denominator.
The leading coefficients of the numerator and denominator are 5 and 2, respectively. Therefore,
the equation of the horizontal asymptote is $y=\frac{5}{2}$.
