Question

Write a function with the following characteristics:
A vertical asymptote at x=-1
An oblique asymptote at y=x+2

Answer 1

for there to be a oblique asymptote the degree of numerator needs to be one higher than the degree of the denominator

Answer 2

you also want the bottom to contain the factor (x+1) since you want a vertical asymptote at x=-1

Answer 3

so let's assume the bottom is x+1
then the top being one degree higher would be of the form ax^2+bx+c

Answer 4

\[\frac{ax^2+bx+c}{x+1}=x+2\]

Answer 5

find a,b, and c

Answer 6

and actually we still have a problem

Answer 7

Okay wow that was a lot thank you but I don't know what a, b, and c would be given the information that I have

Answer 8

that would give us a hole at x=-1 not a vertical asymptote

Answer 9

Okay so what would this problem look like?

Answer 10

@freckles

Answer 11

\[\frac{ax^2+bx+c}{x+1}=x+2+\frac{R}{x+1}\]
there assuming R is not 0 that means x+1 is not a factor of ax^2+bx+c
problem gotten rid of

Answer 12

okay so that would be the answer then? @freckles

Answer 13

no you have to solve for a,b, and c

Answer 14

I still don't know what a, b, and c are.

Answer 15

have you tried multiplying both sides by (x+1)

Answer 16

and expanding
and then comparing both sides
that is one way

Answer 17

can you show me how far you gotten with this?

Answer 18

Well Im basically don't have anything so I can't show you how far I am

Answer 19

so you haven't tried doing any of the things I asked you to?

Answer 20

Im in the middle of doing it just its not coming out right so I'm having to redo it again

Answer 21

can you show me what you have after multiplying both sides of that equation above
the equation \[\frac{ax^2+bx+c}{x+1}=x+2+\frac{R}{x+1}\]
by (x+1)
this was the first step I asked you to do above

Answer 22

\[\frac{P(x)}{x+1}=Q(x)+\frac{R}{x+1} \\ \text{ multiply both sides by } x+1 \\ P(x)=Q(x)(x+1)+R\]
then expand the Q(x)(x+1) part in your problem
Q(x) being (x+2) here

Answer 23

Okay so I don't know what you are doing. I did not learn this in my unit so I think that is why I am so confused at what you are saying.

Answer 24

oh you doing know how to multiply (x+1) on both sides?

Answer 25

\[(x+1) \cdot \frac{ax^2+bx+c}{x+1}=(x+1) [x+2+\frac{R}{x+1}] \\ (ax^2+bx+c) \cdot \frac{x+1}{x+1}=(x+1)(x+2)+(x+1) \cdot \frac{R}{x+1} \text{ by distributive property } \\ (ax^2+bx+c) \cdot 1 =(x+1)(x+2) +R \cdot \frac{x+1}{x+1} \\ ax^2+bx+c=(x+1)(x+2)+R \cdot 1 \\ ax^2+bx+c=(x+1)(x+2)+R\]

Answer 26

expand the (x+1)(x+2)

Answer 27

and then compare both sides to find what a,b, and c are

Answer 28

you will see there are infinitely many choices for c

Answer 29

have you expanded (x+1)(x+2)?

Answer 30

to multiply (x+1)(x+2)
just use distributive property twice
(x+1)(x+2)
=x(x+2)+1(x+2)
=....

Answer 31

x^2+3x+2

Answer 32

great
so you have
\[ax^2+bx+c=x^2+3x+2+R\]

Answer 33

compare both sides
a=?
b=?
c=?

Answer 34

a= 1x^2
b=3x
c=2

Answer 35

well no that means you have
x^2*x^2+3x*x+2=x^2+3x+2+R which is not true

Answer 36

in order for this equation to hold you need to pick a,b,c such that you have the same about of x^2 on both sides
the same amount of x on both sides
and the constant on both sides

Answer 37

for example ax^2=x^2 when a=1

Answer 38

a=1
b=3
c=2

Answer 39

a and b are correct
but for c you ignored the R

Answer 40

c=2+R

Answer 41

\[ax^2+bx+c=x^2+3x+2+R \\ \implies \\ a=1 \\ b=3 \\ c=2+R\]
where we said R couldn't be 0

Answer 42

so as long as you choose R not to be 0 you are fine in your choice for c

Answer 43

okay so what is the final answer then?

Answer 44

have you chose the value of R you want to use yet?

Answer 45

you have infinitely many numbers to choose from except 0 as I said

Answer 46

oh okay well I got it thank you for your help

Answer 47

\[\frac{ax^2+bx+c}{x+1} \text{ has } \text{ va } x=-1 \text{ and } \text{ oba } y=x+2 \\ \text{ if you \choose } a=1,b=3, c=2+R , R \neq 0\]
so R=1 would work
R=12 or R=-12
R=1/2
whatever just you can't choose R to be 0