# Finding the equation of vertical and horizontal asymptotes

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

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

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

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 $y=0$.

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.

• Oh I thought it meant that 3 was multiplied by either the negative square root of 25 or positive. How do you get that equation though? – user3102740 Oct 25 '14 at 15:27
• The quadratic formula. Given a quadratic $ax^2+bx+c=0$, $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ – Sujaan Kunalan Oct 25 '14 at 15:29
• Ah thanks man! I didn't know about this formula – user3102740 Oct 25 '14 at 15:35