Question

need help:
Consider the function f(x) = x^2-16/x+3. Determine all asymptotes of this function including horizontal, vertical, and oblique (slant).

please check:
What are the vertical asymptotes of the function f(x) = 5x+1/x^2+x-6? --> x can not = 2, x = -3 correct?

What are the zeros of the function f(x) = 3x^2-3x-6/3x-6? --> -2 correct?

Consider the function f(x) = x^2+x-12/x-3. Describe the graph of this function. Include all discontinuities, intercepts, and the basic shape of the graph. --> disc = -4? x int = (3,-4)? y int = 3? basic shape = straight line? correct?

Answer 1

is it some assignment thingy ?

Answer 2

it was, i got them wrong but i know they will be on the test

Answer 3

ok, for first part
f(x) = x^2-16/x+3.
in order to get vertical asymptotic set denominator equal to zero and solve for x.
x+3=0
so can you tell me what is vertical asymptote ?

Answer 4

-3

Answer 5

great ! thats correct

Answer 6

now in order to find the Horizontal asymptote , check the following
1)if the degree of the highest variable in numerator and denominator is same then asymptote is simply the ratio of the coefficients of the largest value
For example
f(x)= (3x^2+8)/(5x^3+9) since both numerator and denominator have same degree so Horizontal asymtote is the ratios of their coefficients y=3/5 is horizontal asymptote..

2) if the degree of numerator is greater than denominator, than there is no Horizontal asymptote For example
f(x)= (5x^3+7)/(3x^2+8) since the degree of numerator is greater ( it is 3 x^3) and that of denominator is less ( it is 2 x^2 ) there is no horizontal asymptote.

can you now tell me Horizontal asymptote of
f(x) = x^2-16/x+3. ???

Answer 7

hmmm (x+4)(x-4)/x+3 id say the denom is lower than num

Answer 8

so none

Answer 9

yes you are right. there was no need of factorization !!!.
no it is not one since the degree of denominator is less so there is no Horizontal asymptotic at all.

Answer 10

i c

Answer 11

when there is no Horizontal asymptote then there is oblique (slant) asymtote .

Answer 12

good deal

Answer 13

sorry divide by x+3
for that you have to perform the Long division.
so divide the (x2-16) by (x+3) and tell me what you get after division.

Answer 14

x+3 / x^2 - 16
x^3-16x +3x^2-48
- x^2-16
x^3-16x +2x-32

Answer 15

what is quotient ?

Answer 16

x^2 - 16

Answer 17

i think you went wrong !!!

Answer 18

look here
http://www.wolframalpha.com/input/?i=Long+division%28x%5E2-16%29%2F%28x%2B3%29&a=*MC.Long+division-_*MathMiscWord-

Answer 19

sorry

Answer 20

click on show steps

Answer 21

very good ty

Answer 22

so -7

Answer 23

-7 is remainder !!! quotient is (x-3)

Answer 24

i understand now

Answer 25

so what ever you get in quotient set that equal to y hence
y=x-3 should be slant asymptotic.

Answer 26

y=3 ?

Answer 27

no
y=x-3 is the slant asymptotic

Answer 28

so va = -3, ha = 7/(x-3), oa = y= x-3

Answer 29

there is no Horizontal asymptote !!! told you already both slant and Horizontal cannot occur at the same time.

Answer 30

yes that was correct

Answer 31

so asymptotic stuff is now clear ??

Answer 32

believe so

Answer 33

:)

Answer 34

ty

Answer 35

you are welcome :)

Answer 36

this link might be helpful .
http://www.purplemath.com/modules/asymtote4.htm

Answer 37

What are the vertical asymptotes of the function f(x) = 5x+1/x^2+x-6? --> x can not = 2, x = -3 correct?

What are the zeros of the function f(x) = 3x^2-3x-6/3x-6? --> -2 correct?

Consider the function f(x) = x^2+x-12/x-3. Describe the graph of this function. Include all discontinuities, intercepts, and the basic shape of the graph. --> disc = -4? x int = (3,-4)? y int = 3? basic shape = straight line? correct?

Answer 38

are those correct

Answer 39

yes the first one is correct
f(x) = 5x+1/x^2+x-6 , x=-3, x=2 they are va.

Answer 40

phew

Answer 41

What are the zeros of the function f(x) = 3x^2-3x-6/3x-6? --> -2 correct?

Answer 42

in order to get zeros of polynomial set f(x)=0
so you need to solve
\[\Large \frac{3x^2-3x-6}{3x-6}=0\]
\[\Large \implies (3x^2-3x-6)=0\]
solve this Quadratic equation now.

Answer 43

can you do this now ???

Answer 44

yes

Answer 45

x(x^2-x-2) = x(x+1)(x-2)

Answer 46

ops 3(x+1)(x-2)

Answer 47

yes
solve
3(x+1)=0
(x-2)=0

Answer 48

zero f (x) = -1

Answer 49

i put 2 originally and got it wrong

Answer 50

there are two zeros because we have solved Quadratic equation.
they are
3(x+1)=0
x-2=0
so
x=-1
x=2
are two zeros.

Answer 51

ty