Question

Help with asymptotes. f(x)=x-2/x^2-2x-2
Hole in the graph=?
V.A.=?
H.A.=?
Slant Asymptote=none

Answer 1

I know how to graph it just need help finding the final equation

Answer 2

I also know that we cant factor the denominator so i made it equal to 0

Answer 3

and i got x=4 and x=-2

Answer 4

but i have not idea what to do know

Answer 5

well can you please check the denominator is

\[x^2 - 2x - 2\]

Answer 6

x = 4 and x = -2 aren't solutions to the denominator...

you will need to use the general quadratic formula to find the vertical asymptotes.

the horizontal asymptote is x = 0

since the degree of the numerator is less than the degree of the denominator

Answer 7

and if you use the graphing calculator below you can see when the asymptotes are

https://www.desmos.com/calculator

Answer 8

wait where is the vertical asymptote on that graph?

Answer 9

thats correct... you have a quadratic equation... so it will have 2 solutions......

that's why were are 2 vertical asymptotes...

Answer 10

the 1st task to finding vertical asymptotes...

1. let the denominator = 0 and solve

2. the solutions are the values that cannot be in the denominator...

3. they are the vertical asymptotes...

Answer 11

well when you graph them.... there is 1 vertical at around - 0.5

and 1 at about 3.5

so find the exact values of the asymptotes, you need to use the

general quadratic formula of completing the square on the denominator

Answer 12

\[f(x)=a _{n}x ^{n}+a _{n -1}x ^{n -1}+a _{n -2}x ^{n -2}+...+a _{1}x +a _{0}\] is a polynomial function of a single variable, where \[a _{0},a _{1},...,a _{n -1},a _{n}\] are real valued coefficients (a[0] is also a constant term) and n belongs to the set of positive integers.

A rational function is a ratio of two polynomial functions. Hence
y = P(x)/Q(x) is a rational function if both P(x) and Q(x) are polynomial functions and Q(x) ≠ 0.

I). Factor, if possible, both P(x) and Q(x) completely.
II). State restrictions on the variables (values of x where Q(x) = 0).
III). Divide factors common to both P(x) and Q(x), to simplify.

If there are any factors in the denominator that completely disappear after dividing like factors, then the zeros of those factors indicate where there are removable discontinuities (I guess you can call them holes in the graph, although I don't like that).

If there are any factors that still remain in the denominator, the zeros of these factors indicate infinite discontinuities (vertical asymptotes). Graph of a rational function never intercepts a vertical asymptote. The function tends to either "+ve" or "-ve" ∞ as x "approaches" the vertical asymptote from either the left or the right.

If the degree of P(x) < the degree of Q(x), then there is a horizontal asymptote at y= 0 (the x-axis).
If the degree of P(x) = the degree of Q(x), then there is a horizontal asymptote at y= (leading coefficient of the numerator)/(leading coefficient of the denominator). In other words, the quotient of the leading coefficients.
If the degree of P(x) > the degree of Q(x), then a horizontal asymptote is no longer in existence, and makes way for an oblique asymptote. In other words, an oblique asymptote exists iff the degree of P(x) > the degree of Q(x). This should imply that a horizontal asymptote and an oblique asymptote never co-exist on the graph of a rational function.
the graph of a rational function is allowed to intercept a horizontal asymptote where the function equals the horizontal asymptote.

If the degree of P(x) is greater than the degree of Q(x) by 1, then we have an oblique asymptote which is on a linear diagonal run
\[y =a _{1}x +a _{0}\]. You call this a slant asymptote (I don't like the word slant, either).
If the degree of P(x) is greater than the degree of Q(x) by 2, then we have an oblique asymptote which is a quadratic polynomial
\[y =a _{2}x ^{2}+a _{1}x +a _{0}\].
If the degree of P(x) is greater than the degree of Q(x) by 3, then we have an oblique asymptote which is a cubic polynomial
\[y =a _{3}x ^{3}+a _{2}x ^{2}+a ^{1}x +a _{0}\]...
In other words, oblique asymptotes are in fact, imaginary polynomial functions whose equations can be determined by long division. I suggest dividing the simplified function, for the ease of it. Also, like the horizontal asymptote, the graph of the rational function intercepts the oblique asymptote where the remainder is zero.

Now, @Cool1003 you can apply this lesson on Rational Functions, to solve other similar problems.

GOOD LUCK!!

Answer 13

umm, thats a bit confusing

Answer 14

I mean i understand horizontal asymptote, but the vertical one is still make no sense

Answer 15

OK, your question is \[f(x)=\frac{ x -2 }{ x ^{2}-2x -2 }\]Correct?

Answer 16

ok... so one of the basics of a fraction is that you can't have a zero denominator

is that ok...?

Answer 17

yes i know that i ment just in this particular question its hard to understand

Answer 18

i know that if the denominator was x-3 it would be x=3

Answer 19

because you cant divide by a 0

Answer 20

ok... so you have a quadratic equation.... and every quadratic has 2 solutions...

sometimes 2 different... 2 the same or 2 complex

so you need to find the values of x that make the denominator zero

so you have

\[x^2 -2x - 2 = 0\]

this quadratic equation can't be solved by factoring...

it can be solved by using the general quadratic formula

\[x = \frac{-b \pm \sqrt{b^2- 4ac}}{2a}\]

or by using the complete the square method...

the solutions you find... will be the vertical asymptotes...

Answer 21

Ok but how do i know whats my b,a, and c is

Answer 22

ok... for any quadratic equation it is normally written as

\[ax^2 + bx + c \]

so match a, b and c to your quadratic....

Answer 23

what do you think the values of a, b and c are...?

Answer 24

\[ax^2 + bx + c\]
\[x^2 -2x - 2\]

Answer 25

ok a=1, b=-2, c=-2

Answer 26

great... substitute them and evaluate, then you get the 2 values of x, that are the vertical asymptotes...

I do find it a bit unusual that you have working on these questions and don't know the General Quadratic Formula.

Answer 27

I got \[2\pm \sqrt{-16}/2\]

Answer 28

not sure if i did the + and - correctly though

Answer 29

well seems a slight error

\[x = \frac{2 \pm \sqrt{(-2)^2 - 4 \times 1 \times -2}}{2 \times 1} = \frac{2 \pm 2\sqrt{3}}{2}\]

so the asymptotes exist at

\[x = 1 \pm \sqrt{3}\]

hope it helps

Answer 30

ah i forgot to square the -2

Answer 31

and is that is or do we have to do something else with the x=1±√3

Answer 32

that's it for the vertical asymptotes.... I'd leave them as exact values

you may write it as

\[x = 1 + \sqrt{3}...and... x = 1 - \sqrt{3}\]

Answer 33

when you graphed the equation... the the x axis was a horizontal asymptote

this is because the degree of the numerator is 1

and the degree of the denominator is 2

when you see degree of the numerator less than the degree of the denominator then the asymptote will be x = 0

Answer 34

so its x-2/x=1±√3

Answer 35

no... the equation doesn't change for

\[f(x) = \frac{x -2}{x^2 -2x -2}\]

the vertical asymptotes exist at

\[x = 1 - \sqrt{3} ...and.... x = 1 + \sqrt{3}\]

the horizontal asymptote is

x = 0

there are no slant asymptotes.

Answer 36

ah ok