Given rational function:
[tex]Q(x)\text{ = }\frac{5x^2-17x-12}{3x^2-10x-8}[/tex]Vertical asymptote
To find the vertical asymptote, we set the denominator function to zero and solve for x
First, we factor the function.
[tex]\begin{gathered} Q(x)\text{ =}\frac{\left(5x+3\right)\left(x-4\right)}{\left(3x+2\right)\left(x-4\right)} \\ \end{gathered}[/tex]Since we can cancel (x-4) is present in the numerator and denominator, x= 4 is a hole
Setting the other factor in the denominator to zero:
[tex]\begin{gathered} 3x\text{ + 2 = 0} \\ 3x\text{ = -2} \\ Divide\text{ both sides by 3} \\ \text{x = -}\frac{2}{3} \end{gathered}[/tex]Hence, the vertical asymptote is x =-2/3
Horizontal asymptote
The first step is to compare the degrees of the numerator and denominator function. Since, the degrees are equal, we divide the leading co-efficients
Recall that the leading co-efficient is the coefficient of the term with the highest degree of the polynomial.
Hence, we have :
[tex]\begin{gathered} y\text{ = }\frac{5}{3} \\ 5\text{ is the leading coeffient of the numerator function and 3 is the leading coefficient of the } \\ denominator\text{ function} \end{gathered}[/tex]Hence, the horizontal asymptote is y = 5/3
Oblique asymptote
Oblique asymptotes occur when the degree of the denominator of a rational function is one less than the degree of the numerator.
Hence, there is no oblique asymptote
