Answer
Graph the function as:
Work Step by Step
Step 1: Substitute $x=-x$.
$\begin{align}
& f\left( x \right)=\frac{3{{x}^{2}}+x-4}{2{{x}^{2}}-5x} \\
& f\left( -x \right)=\frac{3{{\left( -x \right)}^{2}}+\left( -x \right)-4}{2{{\left( -x \right)}^{2}}-5\left( -x \right)} \\
& =\frac{3{{x}^{2}}-x-4}{2{{x}^{2}}+5x}
\end{align}$
Therefore, the function $f\left( -x \right)$ is not equal to $-f\left( x \right)$ and the $f\left( x \right)$. So, the graph of the function is not symmetrical about the $y$ axis and the origin.
Step 2: To calculate the x intercepts equate $f\left( x \right)=0$.
$\begin{align}
& \frac{3{{x}^{2}}+x-4}{2{{x}^{2}}-5x}=0 \\
& x=1
\end{align}$
Step 3: To calculate the y intercept evaluate $f\left( 0 \right)$.
$f\left( 0 \right)=\infty $
Thus, the y-axis is an asymptote.
Step 4: Since the degree of the numerator is equal to the denominator, the horizontal asymptote is:
$y=\frac{3}{2}$
Step 5: For the vertical asymptote, equate the denominator to 0.
$x=0,\frac{5}{2}$
