Question

Graphing an equation/function. f(x)= (-6)/(x+3)
See attachment. @Calculus1

Answer 1

Have you taken the derivative?

Answer 2

Yes, (6)/(x+3)^2

Answer 3

Have you set equal to zero, to find any relative extrema?

Answer 4

No, I got lost on that part. Would we use -3?

Answer 5

You're thinking. Actually, we can't get it to equal 0, but at x=-3, it is undefined. So x=-3 is a critical point.

Answer 6

More precisely x=-3 is a vertical asymptote.

Answer 7

And we have no local extrema.

Answer 8

Horizontal asymptotes would be 0, right?

Answer 9

at*

Answer 10

If you put a value in for x<3 into the first derivative, do you get a positive or a negative answer?

Answer 11

Positive

Answer 12

Yes, the horizontal is at y=0.

Answer 13

What does the positive value for the derivative tell you about the function?

Answer 14

It's increasing.

Answer 15

Correct. You should also check to see what the function is doing for x>-3.

Answer 16

It's increasing. o.o

Answer 17

Great. Have you looked at concavity yet?

Answer 18

Umm, no, got lost on that one two.

Answer 19

Wait is it possible to be increasing on bot sides?

Answer 20

both*

Answer 21

Yes.

Answer 22

If you have a fractional equation, such as (x+2)/(2x^2 + 3) = 0 and it is set equal to zero, you can ignore the denominator and solve the new equation x+2=0.

Answer 23

Umm, the second derivative is (-12)/(x+3)^3 how do you find the inflection point?

Answer 24

Does it have any?

Answer 25

Since the second derivative is -12/(x+3)^3 and we cannot have -12=0, so there are no inflection points.

Answer 26

So I got this, it's increasing on [-inf,-3] and on [-3,inf] and it's not decreasing.

Answer 27

Those should be parenthesis instead of brackets.

Answer 28

aaah okay, thanks! I think that's it. Thank you so much.

Answer 29

(-inf,-3) and on (-3,inf)

Answer 30

You're welcome.