The equations of the vertical asymptotes are the values of the zeroes of the denominator
The given algebraic rational function is
[tex]s(x)=\frac{6x^2+1}{2x^2+3x-2}[/tex]
Equate the denominator by 0 to find its roots
[tex]2x^2+3x-2=0[/tex]
Factor the left side
[tex]\begin{gathered} 2x^2=(2x)(x) \\ -2=(-1)(2) \\ (2x)(2)+(x)(-1)=4x-x=3x \\ 2x^2+3x-2=(2x-1)(x+2) \\ (2x-1)(x+2)=0 \end{gathered}[/tex]
Equate each factor by 0
[tex]\begin{gathered} 2x-1=0 \\ 2x-1+1=0+1 \\ 2x=1 \\ \frac{2x}{2}=\frac{1}{2} \\ x=\frac{1}{2} \end{gathered}[/tex][tex]\begin{gathered} x+2=0 \\ x+2-2=0-2 \\ x=-2 \end{gathered}[/tex]
Then the equations of the vertical asymptotes are
[tex]\begin{gathered} x=-2\rightarrow smaller \\ \\ x=\frac{1}{2}\rightarrow larger \end{gathered}[/tex]
To find the horizontal asymptote
Since the greatest power of up and down is 2
Then divide the coefficients of x^2 up by down
Since the coefficient of x^2 up is 6
Since the coefficient of x^2 down is 2
Since 6/2 = 3
Then the equation of the horizontal asymptote is
[tex]y=3[/tex]
