Finding the equation of vertical and horizontal asymptotes

Question

I am having some trouble understanding these two questions. Any help is appreciated. Scanned questions are included at the end.

6) We are given the function $ f(x) =\frac{1 - 2x} {2x^2 - 3x - 2} $

6 a) Find the equation of the vertical asymptotes. Explain how.

For the above question how did they first get the equation $ x =( 3 \pm √25 ) / 4, $

and then get x = 2 and x = -1/2 out of it?

6 b) Find the equation of the horizontal asymptotes. Use a limit.

For this question I understand that when the degree of the numerator is less than the degree of the denominator it results in a horizontal asymptote. Thus here we get y = 0. Right? But I would still like to know if it is the same procedure they used in the answer sheet to get the answer 0/2.

Answer 1

For question a, the quadratic formula is used to solve the quadratic equation.

$$2x^2-3x-2=0$$

$$x_{1,2}=\frac{3\pm\sqrt{(-3)^2-4(2)(-2)}}{2(2)}=\frac{3\pm5}{4}=-\frac12,2$$

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For question b, consider $\dfrac{\frac{1}{x^2}-\frac2x}{2-\frac3x-\frac{2}{x^2}}$ as $x\to\infty$. The fractions become $0$ and the limit is $\frac02=0$. Hence the horizontal asymptote is $y=0$.

Answer 2

For part $a)$, $\dfrac{3\pm\sqrt{25}}{4}=\dfrac{3\pm 5}{4}$ which equals either $\dfrac{3+5}{4}$ or $\dfrac{3-5}{4}$, giving us $x=2$, and $x=-\frac{1}{2}$, respectively.