#1**+2 **

a)

\(lim_{x\rightarrow-\infty}\frac{3x^2+14x-5}{x^2-25}\\ =lim_{x\rightarrow-\infty}\frac{x^2(3+\frac{14}{x}-\frac{5}{x^2})}{x^2(1-\frac{25}{x^2})}\\ =lim_{x\rightarrow-\infty}\frac{3+\frac{14}{x}-\frac{5}{x^2}}{1-\frac{25}{x^2}}\\ =lim_{x\rightarrow-\infty}\frac{3+0-0}{1-0}=3\)

b)

horizontal asymptotes happen when the limit as x approaches infinity or negative infinity approaches some number

as found above, **a horizontal asymptote is y=3**

you can also check the limit as x approaches positive infinity as sometimes there are 2 horizontal asymptotes, but in this case that limit also equals 3, so there's only one horizontal asymptote.

c)

\(\frac{3x^2+14x-5}{x^2-25}=\frac{(x+5)(3x-1)}{(x-5)(x+5)}\\ \)

first step is to factor the numerator and denominator

then you see that x+5 can be canceled out. this means that there's a hole (removeable discontinuity) at x=-5

after canceling out you're left with:

\(\frac{3x-1}{x-5}\)

to find the coordinates of the hole plug in x=-5 into that and you get y=8/5, so **the hole is at (-5,8/5)**

to find vertical asymptotes, use the equation that's left over after the canceling and set the denominator equal to 0

x-5 = 0, x = 5

**A vertical asymptote is x=5**

Guest Apr 21, 2020

#2**+1 **

Oh Wow! Thank You so much! This all looks so much understandable now. Do you mind doing the same here?

https://web2.0calc.com/questions/let-g-x-x-4-x-2-x-12

GAMEMASTERX40
Apr 21, 2020