# Thread: Asymptotes and intercepts: help checking work

1. ## Asymptotes and intercepts: help checking work

I had to find the vertical and horizontal asymptotes as well as the x an y intercepts..could someone please help me by checking my work...there are only 8 problems...you do not have to check them all, but I wouldreally appreciate the help because I do not know if I am doing these correctly....thank you

2. Hello, aikenfan!

You did really good work . . . with a few slips here and there.

$12)\;\;f(x)\:=\:-\frac{x+2}{x+4}$
That minus-sign is in front of the fraction . . .

For the x-intercept, we have: . $-(x +2)\:=\:0 \quad\Rightarrow\quad x + 2 \:=\:0$

. . $x \:=\:-2\quad\Rightarrow\quad (-2,0)$

$18)\;\;f(x)\:=\:\frac{x^2-25}{x^2+5x}$
VA: . $x^2+5x \:=\:0 \quad\Rightarrow\quad x(x+5)\:=\:0\quad\Rightarrow\quad x \:=\:0,-5$

Vertical asymptotes: . $x = 0,\;x = -5$

$43)\;\;f(x) \:=\:\frac{2x^2+1}{x}$
x-intercept: . $2x^2+1\:=\:0\quad\Rightarrow\quad x^2 \:=\:-\frac{1}{2}\quad\Rightarrow\quad x \:=\:\sqrt{-\frac{1}{2}}$ . . . not a real number.

There is no x-intercept.

$44)\;\;g(x) \:=\:\frac{1-x^2}{x}$
x-intercepts: . $1-x^2\:=\:0\quad\Rightarrow\quad x^2 \:=\:1\quad\Rightarrow\quad x \:=\:\pm1\quad\Rightarrow\quad(1,0),\;(-1,0)$

$45)\;\;h(x) \:=\:\frac{x^2}{x-1}$
y-intercept: . $y \:=\:\frac{0^2}{0-1} \:=\:0\quad\Rightarrow\quad (0,\,0)$

3. Thank you very much for all of your help...I just have one more quick question, if you don't mind...for number 18, how do i figure out the horizontal asymptote? I think it would be 1, but I'm not sure.

4. Hello again, aikenfan!

For number 18, how do i figure out the horizontal asymptote?
Yes, the horizontal asymptote is: $y = 1$

We want: . $\lim_{x\to\infty}\frac{x^2-25}{x^2+5x}$

Divide top and bottom by $x^2$: . $\lim_{x\to\infty}\frac{\frac{x^2}{x^2} - \frac{25}{x^2}}{\frac{x^2}{x^2} + \frac{5x}{x^2}}$ . **

Then: . $\lim_{x\to\infty}\frac{1 - \frac{25}{x^2}}{1 + \frac{5}{x}} \;=\;\frac{1-0}{1+0} \;=\;1$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

Rule: Divide top and bottom by the highest power of $x$
. . . . in the denominator.

5. Note that we should find $\lim_{x \to - \infty} \frac {x^2 - 25}{x^2 + 5}$ as well. But in this case, it would give the same value, so that's fine