Question

4. What are the points of discontinuity? Are they all removable? Please show your work.

Answer 1

factor the denominator, then set it equal to zero and solve
those are the discontinuities

Answer 2

if, once you factor, you can also cancel, that will remove the discontinuity

Answer 3

y = (x-5)/ x^2-6x+5

Answer 4

can you help me step by step please

Answer 5

factor the denominator

Answer 6

\[y=\frac{x-5}{(x-5)(x-1)}\]

Answer 7

the denominatior for my quation is x^2-6x+5 ight

Answer 8

then if
\[(x-5)(x-1)=0\] you have
\[x=5,x=1\] and those are your discontinuities because you cannot divide by zero

Answer 9

yes the bottom is the denominator

Answer 10

okay can we got slow step by step like u ask me and i answer setc back and forth likethat please
im having a hard time understanding and i want to learn

Answer 11

how to i facto out that denominator is confusing to me im sorry :(

Answer 12

ok
do you know how to factor
\[x^2-6x+5\]? because that is what you have to do next

Answer 13

you do not "factor it out" you just factor it in to a product, as if you were solving the equation
\[x^2-6x+5=0\]

Answer 14

okay im not sure how to do that

Answer 15

think of two numbers that, when multiplied together give \(5\) and when added together give \(-6\)

Answer 16

2.5 x 2.5 = 5 right

Answer 17

oh no
think of two integers, not decimals

Answer 18

also \[2.5\times 2.5\neq 5\]

Answer 19

there are really only two pairs of integers that multiply to give \(5\)

Answer 20

this is so hard lol im sorry for needing help

Answer 21

no problem

Answer 22

it is clear that \(1\times 5=5\) ?

Answer 23

yes

Answer 24

and also \(-1\times -5=5\) right?

Answer 25

may i ask why there are 3 equations like that ? and how each gets smaller

Answer 26

yes that is correct

Answer 27

but \(1+5=6\) and \(-1-5=-6\) so you have to pick \(-1,-5\) to factor

Answer 28

okay so factor -1 and -5?

Answer 29

why is there the 3 copies on the equations im confused by that

Answer 30

factor as
\[x^2-6x+5=(x-1)(x-5)\]

Answer 31

if you multiply out on the right, you get the quadratic on the left

Answer 32

there are not three copies, not sure what you mean by that

Answer 33

ill screen shot for u

Answer 34

ok

Answer 35

well idk how to on here so. is it just supposed to be x^2-6x+5 = (x-1)(x-5)

Answer 36

yes those are equal

Answer 37

okay good

Answer 38

so
\[\frac{x-5}{x^2-6x+5}=\frac{x-5}{(x-5)(x-1)}\]

Answer 39

because the bottoms are equal

Answer 40

now we are almost done

Answer 41

you are looking for the "discontinuities" right?

Answer 42

okay i understand

Answer 43

yes the points of discontinuties and if they are removable

Answer 44

you cannot divide by zero
set
\[(x-5)(x-1)=0\] and solve for \(x\)

Answer 45

do you know how to do that?

Answer 46

would it b 5-5 amd 1-1?

Answer 47

it would be \(x=5\) or \(x=1\) so yeah, sort of

Answer 48

\[x-5=0\\
x=5\] or
\[x-1=0\\
x=1\]

Answer 49

those are your discontinuities, \(5,1\)

Answer 50

okay so 5,1 are my discontinuties

Answer 51

yup
now for the removable part

Answer 52

okay

Answer 53

we have
\[y=\frac{x-5}{(x-5)(x-1)}\] so you can cancel right?

Answer 54

yes

Answer 55

what do you get when you cancel?

Answer 56

0?

Answer 57

hmm no

Answer 58

can you explain it to me please

Answer 59

\[y=\frac{x-5}{(x-5)(x-1)}=\frac{\cancel{x-5}}{\cancel{(x-5)}(x-1)}=\frac{1}{x-1}\]

Answer 60

so x-1?

Answer 61

or 1/x-1

Answer 62

\[\frac{1}{x-1}\]

Answer 63

okay great now what

Answer 64

we "removed" the \(x-5\) right?

Answer 65

correct

Answer 66

so \(5\) is the "removable discontinuity"

Answer 67

and 1 is not

Answer 68

okay that makes sense, does the complete the answer?

Answer 69

finally, yes

Answer 70

okay, i appreciate the help so much. I have a few more ill be posting in a bit. hopefully your on. i love learning and just not getting an answer

Answer 71

no problem