Question

Help please

Answer 1

i already got the asymptotes

Answer 2

if you'd like me to check the asymptotes I'd be ok with that

Answer 3

Okay so I have
26. vertical x=0
no horizontal
27. vertical x=-4
horizontal y=0
28. vertical x=-2.748
horizontal y=2
29 vertical x=-3
horizontal y=1

Answer 4

I can't see the problem for 29

Answer 5

asymptotes look good

for x-intercepts simply take the factors from the numerators and set them equal to zero and find the xvalues

Answer 6

so start by factoring x^2- 12x + 20

Answer 7

26. x intercept (10,0)(2,0)
y intercept none
27. x intercept (8,0)(-1,0)
y intercept (0,-1/8)
28. x intercept (1,0)(-1-sqrt3/2 , 0) (-1+sqrt3/2 , 0)
y intercept (0,1/7)
29. x intercept (1,0)
y intercept (0,-1/3)

Answer 8

sorry it took me a while

Answer 9

good, well done

Answer 10

for removable discontinuities you will have to factor the numerator and the denominator and see which produce roots that can be cancelled in the numerator/denominator

Answer 11

okay so

Answer 12

afaik 26 and 27 don't have any removable discontinuities so start with 28

Answer 13

okay so do i factor the num. and denom. separately

Answer 14

yes

Answer 15

then re-write the function in factored form and see if anything cancels out

Answer 16

okay so for example 28 it would be (x-1)(2x^2-2x-1)

Answer 17

I factored the numerator

Answer 18

yeah, I don't think 28 has any removable discontinuities

29 does have one though

Answer 19

um is it infinite discontinuity x=0?

Answer 20

when you factor the numerator and denominator of 29 what do you get?

Answer 21

(x-1)(x+5) and (x+3)(x+5)

Answer 22

so x+5 and cancels out

Answer 23

good, so x = - 5 is the removable discontinuity

that's it

Answer 24

thanks !