Question

give the vaules of all the vertical asymptotes,horizonal asympyotes,slant asymptotes, and the holes to the equation (2x-6) / (x^2 - 3x)

Answer 1

factor the top and bottom so that you can cancel out like parts

Answer 2

how do u do that i do not understand it period

Answer 3

if you cannot factor a polynomial yet, then you are not ready to do this kind of problem

Answer 4

i have to its my class work in percal

Answer 5

then you have to learn how to factor polynomials first, or you will not be able to work out the maths the build on it.

Answer 6

2x-6

do 2x and 6 have any common factors?

Answer 7

yes 12

Answer 8

12 would be a common multiple; not a common factor

the factors of:

2x: 2,x
6: 2,3

notice that the common factor that they share is "2"

Answer 9

wats the difference between a common factor and a common multiple and yes i see that

Answer 10

factors are the smaller parts that make up a number; a multiple is what you get when you multiply the numbers together.

they are like 2 sides of the same coin. different, but related.

Answer 11

oh okay

Answer 12

\[2x-6 = 2(x) - 2(3)\]
\[2(x) - 2(3)=2(x-3)\]this is the "factored" form for the top

Answer 13

lets factor the bottom

x^2 - 3x

the factors of:

x^2: x, x
3x: 3, x

notice that the factor that they have in common is "x"

Answer 14

I recommend a quick refresher when you get a chance, @lissia_liverman
http://themathpage.com/aPreCalc/algebraPre.htm

Answer 15

thanks CliffSedge

Answer 16

\[x^2 - 3x=x(x)-x(3)\]
\[x(x)-x(3)=x(x-3)\]is the factored form of the bottom, giving us

\[\frac{2x-3}{x^2-3x}~=~\frac{2(x-3)}{x(x-3)}\]

Answer 17

with this in the factored form, we can start to recognize the holes and vertical asympotes

Answer 18

u no ur good at explaining things amistre64

Answer 19

lol, im just too old to know better :)

Answer 20

lol nice i like that :)

Answer 21

"holes" are what happens when we can remove things. notice that the like factors top to bottom can be "removed"\[\frac{2\cancel{(x-3)}}{x\cancel{(x-3)}}\]
therefore when x-3 = 0 we have a "hole"
the vertical asymptotes are what we are left with that zero out the bottom
\[\frac{2}{x} \]in this case, when x=0 we get a zero in the bottom and a vertical asymptote

Answer 22

the horizontal and slant asymptotes are what happens when the values of x get very big or very small.

Answer 23

\[\frac{2x-6}{x^2 - 3x}\]when x get very big (or very small) the leading terms take control of the setup. for example:\[\frac{2(1,000,000)-6}{(1,000,000)^2-3(1,000,000)}\]is "close to"
\[\frac{2(1,000,000)}{(1,000,000)^2}\]
the other parts dont contribute enough to the setup to matter at these large values

Answer 24

oh okay how would you graph this original problem

Answer 25

so, to find the horizontal (or slant) we focus on simplifying the leading terms only and see how they act when x grows into infinity

\[\frac{2x...}{x^2...}\to\ \frac{2}{x}\]
as x gets larger and larger, this setup gets closer and closer to zero. Which will be the horizontal asymptote then

graphing it? that might be a bit complicated by hand.

Answer 26

i can determine a sketch of it using the information from the stuff we have already found to make a close approximation to how it should look

Answer 27

thank u

Answer 28

and then to fill in the way it cozies up to the VA and the HA

Answer 29

\[\frac{2(x-3)}{x(x-3)}\]
lets test a value of x that is clearly between 0 and 3, say x=1. And we are interested in the sign of the answer rather than the value itself, the sign tells us what side of the x axis we want to be on
\[\frac{2(1-3)}{1(1-3)}\]
\[\frac{+(-)}{+(-)}=+\]
everything between 0 and 3 will be positive; lets use x=4 since it is on the other side of the 3
\[\frac{2(4-3)}{4(4-3)}\]
\[\frac{+(+)}{+(+)}=+\]
And x=-4 is good enough for the other side of 0
\[\frac{2(-4-3)}{-4(-4-3)}\]
\[\frac{+(-)}{-(-)}=-\]



so something similar to this