## Precalculus (6th Edition) Blitzer

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. $x=0,\frac{5}{2}$