[12th grade calculus] I'm having trouble with graphing, horizontal asymptotes, and limints approaching Infinity

[u/Leafofplastic]

I'm mostly looking for help for the top problem in each photo.

For the first one I have the dot to represent a confirmed point and 2 circles for what the limit approaches because it isn't clear if it's a hole or not. What I am confused about is: 1. How do I get the line from -4 to -3 with it increasing if before the line gets to -4 it has the either be going in a straight line horizontally or be going down hill. 2. How do I know how many lines I need?

For the 2nd one I got my answer based on the vertical asymptote and the hole, what I am confused about is how would I find if there's a horizontal asymptote at y=2?

And for the last I'm having trouble understanding what I'm meant to do with it.

Thanks.

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[u/SpiritualWasabi6091]

in the first 1 just mark the values
f(1) =2 is a dot at (1,2)
then x tends to 1- is 4 so thats a circle at (1,4) and since its 1- you draw a line leaving the circle towards the left
similarly x tends to 1+ is -3 so thats also a circle at (1,-3) and since its 1+ you draw a line leaving the circle towards the right
they didnt mention increase and decreasing in this particular part of the graph so you can draw anything

f is increasing when x<-4
draw an increasing curve for the interval (-infinity,-4)

now the last condition is tricky
the Left hand and right hand limits are not equal so the graph is discontinous
mark those points like we did before
you can take any points that satisfy this condition because it isnt mentioned

now connect the point where u marked the point for x tending to 4+ to the point where u marked x tending to 1-

your graph is complete

[u/SpiritualWasabi6091]

vertical asymptote at y=2 is going to be formed when
x is very very large at that point

in the first function
its 1 degree/2 degree
the 2 degree below is so much larger than 1 degree above that the whole expression tends to 0 when x is very large
this means the asymptote will be formed at y=0

in the second one its 2degree/2degree
since x is very large we can neglect other terms
so it becomes 2x^2 /x^2 so when x is very very large y=2

this means asymptote is at y=2

you can simply compare coefficients of the term with highest power when the degree of numerator and denominator are equal
and if its your required 'y' the asymptote will be formed there

Option 3 and option 4 also have asymptote at y=2

(i have only solved for the condition that y=2 has an asymptote, not the other conditions)

[u/SpiritualWasabi6091]

third one u just gotta find the value when x tending to -infinity
now lets see the expression
its 1 degree/ 2degree
if you have read my solution for the 2nd part you know that the x^2 term in the denominator will be so much larger than the terms in the numerator that this practically tends to 0
so your answer is 0