Question

3+x is the divisor and 5x^2 is the dividend; I'm trying to find the slanted asymptote using long division

Answer 1

then you will want to disregard the remainder most likely

Answer 2

Yes I was told to disregard it

Answer 3

can you setup the division?

Answer 4

I'm not sure on what to do

Answer 5

setting it up to start with would be prudent. the rest is just division the same way that we always do long hand division

Answer 6

the parts of a division problem should be well known by now. where to place the divisor and dividend are elementary school stuff.

Answer 7

the long hand process itself is also elementary school, its just a matter of working the process that we learned way back when, but this time its not with constants

Answer 8

Do I need to add in x's in the dividend to make this work?

Answer 9

you dont need to, but some texts feel that it might help you keep track of things

Answer 10

if you were doing synthetic division, the add ins are extremely helpful .... but for longhand its just not all that critical

Answer 11

dividend (\(\div\)) divisor, therefore we setup as


-------------
x+3 | 5x^2


getting them into a 'proper' polynomial format is useful, which is why the x+3 and not 3+x

now, how many times does x go into 5x^2?

Answer 12

Twice

Answer 13

not quite:

x * 5x = 5x^2

so x goes into 5x^2, 5x times

Answer 14

5x
-------------
x+3 | 5x^2

now we multiply and subtract the results .... just like we do when they are numbers


5x
-------------
x+3 | 5x^2
-(5x^2+15x)
------------
-15x

now we repeat, how many times does x go into -15x?

Answer 15

15?

Answer 16

x, times (-15) = -15x but you were close :)

Answer 17

now multiply and subtract ....

5x - 15
-------------
x+3 | 5x^2
-(5x^2+15x)
------------
-15x
-(-15x-45)
-----------
45

what do you propose we do next?

Answer 18

Disregard the remainder and that's my slanted asymptote

Answer 19

thats correct, but how do we know we have reached the 'remainder' ?

Answer 20

Because there's nothing to drop down from the dividend?

Answer 21

not quite accurate.

when the part we have left over is of a lesser degree than what we are using to divide, then all the rest of the process is just working the remainder

Answer 22

45 <- degree 0
----
x+3 <- degree 1

Answer 23

so yes, we can disregard any further processing :) and that leaves us with the slant asymptote

Answer 24

@amistre64 am I allowed to divide the sides by five or take out five to make it easier for me to graph?

Answer 25

the 5 is part of the asymptote ... for example: the graph of y=3x is not the same the graph for y=x

Answer 26

5(x-3) is the asmptote line ..... we simply cant alter it

Answer 27

I know how to do the -15 but how do I graph the 5x piece or where does the 5x piece start on the graph

Answer 28

you have a slope and an y intercept .... plot the intercept, and then step out the slope

Answer 29

or, let x=0 to define one point, then let x=1 to define another point .... and draw the line between them

Answer 30

or, let x=0 to define one point, and then let y=0 to define the other

0 = 5x-15, solve the x intercept

Answer 31

x=3