Question

PLEASE HELP!!!!! Is there a hole in the equation
(x^2-3x+2)/(x-2) if there is what is it?

Answer 1

factor and cancel i guess

Answer 2

please help me @satellite73

Answer 3

if you can cancel the common factor of \(x-2\) then there will be a hole at \(x=2\) and \(y=\) whatever you get when you replace x by 3

Answer 4

did you factor?

Answer 5

hint, i cannot be hard, because you know one of the factors is \(x-2\) in the numerator

Answer 6

when i factored i got (x-1)(x-2)

Answer 7

ok good
now cancel

Answer 8

\[\frac{(x-1)(x-2)}{x-2}=?\]

Answer 9

so i'm left with (x-1)/1 which is x-1

Answer 10

yes it is

Answer 11

and thats all i have to do?

Answer 12

hold on i am totally wrong

Answer 13

yeah...

Answer 14

its (2,1)

Answer 15

thanks anyway

Answer 16

when you cancel \(x-2\) you are getting rid of the zero in the denominator at \(x=2\)
if you replace \(x\) by \(2\) you get \((2,1)\)

Answer 17

can you help me with another?

Answer 18

sure

Answer 19

(5x^2+10x)/(-4x-8)

Answer 20

and can you help me find the vertical asymptote too?

Answer 21

factor a common factor of \(5x\) out of the numerator and \(-4\) out of the denominatorq

Answer 22

so i get 5x(x+2)/-4(x+2)

Answer 23

\[\frac{5x^2+10x}{-4x-8}=\frac{5x(x+2)}{-4(x+2)}\]

Answer 24

right

Answer 25

leaves you with only \[-\frac{5}{4}x\]

Answer 26

so i get the hole at (-2,2.5)

Answer 27

no @satellite73 !!!!! why did you leave me?????

Answer 28

lol

Answer 29

i really need help you have NO idea

Answer 30

ok yeah you got it right \((-2,2.5)\)

Answer 31

how would i find the vertical asumptote?

Answer 32

in those two examples there are none

Answer 33

wow, how do you know?

Answer 34

because you were left with no denominators when you factored and cancelled

Answer 35

if you had a denominator left, the vertical asymptote would be the number that would make the denominator zero

Answer 36

wasn't i left with 5x/-4? isnt -4 a denominator than?

Answer 37

ok yes it is a denominator, but it is just a number

Answer 38

and we need a denominator with x in it

Answer 39

yes of course
\[y=-\frac{5}{4}x\] is just a line

Answer 40

if you had
\[y=\frac{x}{x-1}\] then you would have a vertical asymptote at \(x=1\)

Answer 41

one last i found the x and y intercepts to be (0,0) and (0,0) is that correct?

Answer 42

i think i understand the vertical asymptote now

Answer 43

for \(y=-\frac{5}{4}x\) yes

Answer 44

it is a line through the origin with slope \(-\frac{5}{4}\)