1
$\begingroup$

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.

enter image description here

$\endgroup$
0
$\begingroup$

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 $y=0$.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – user3102740 Oct 25 '14 at 15:27
  • $\begingroup$ The quadratic formula. Given a quadratic $ax^2+bx+c=0$, $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ $\endgroup$ – Sujaan Kunalan Oct 25 '14 at 15:29
  • $\begingroup$ Ah thanks man! I didn't know about this formula $\endgroup$ – user3102740 Oct 25 '14 at 15:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.