Question

@phi

Answer 1

can you factor the top and the bottom (and please repost the fraction)

Answer 2

if you factor it you get -5x+6/-1

Answer 3

the bottom is a "difference of squares" a^2 - b^2
it factors into (a+b)(a-b)

Answer 4

of yeah ok I will do that

Answer 5

?/(x+1)(x-1) I have no Idea how to factor the top I tried it another question with that same equation either on the top or bottom

Answer 6

to factor a quadratic, look at the last number
+6
the + means the factors will have the *same sign* (- means different signs)
now list all pairs of numbers that when you multiply them give you 6:
1,6
2,3
(nice short list)

next, look at the number in front of the x: -5
the - means the largest factor is - (+ would mean the largest factor is +)

from the first step, we know both factors have the same sign... both will be -

look at our list with both numbers assigned - and add them. Do any add up to -5 ?

-1-6= -7 no
-2 - 3= -5 yes.
-2 and -3 are our guys. we write

(x -2)(x-3)

Answer 7

ok so how would we graph this

Answer 8

You probably skipped that lesson... but math builds on the earlier stuff... if you don't learn each step you will get stuck

Answer 9

ok you now have
\[ y = \frac{(x-2)(x-3)}{(x+1)(x-1)} \]

the zeros (where y=0 and the curve crosses the x-axis) happens when either of the top terms is zero. what x values cause y to be 0?

Answer 10

uhm -1, 1, 2, and 3 right?

Answer 11

only when the top is 0

Answer 12

ok so then it is 2 and 3

Answer 13

ok, make a note: zeros at x=2 and x=3

can you find the horizontal asymptote ?

Answer 14

uhm idk what to do how would I reduce that fraction

Answer 15

you do the same thing we did on the other problem...or was that your twin working that problem ?

Answer 16

hahaha yeah I know but when I do that the x's should disapear right?

Answer 17

yes, we just get y = # (ratio of the coefficients of the highest order term)

Answer 18

ok but when we get rid of the x's the denominator becomes 0 because of 1-1

Answer 19

using only the high order terms you get
\[ y = \frac{x^2}{x^2} = 1\\ y=1\]

Answer 20

oh yeah ok so the horizontal is 1?

Answer 21

yes y=1

next, what are the vertical asymptotes ? (where you divide by 0)

Answer 22

-5/0 right?

Answer 23

you have
\[ y = \frac{(x-2)(x-3)}{(x-1)(x+1)} \]
what x's cause a divide by 0 ?

Answer 24

bottom

Answer 25

yes, but specifically what x values ?

Answer 26

1 and -1

Answer 27

ok, so what we know is

1 horizontal asymptote y=1
2 vertical asymptotes x= -1 and x=1
2 zeros at x= 2 and x=3

Answer 28

can you plot those lines and/or points ?

Answer 29

it looks like this

Answer 30

now we need to fill in the curve.

Answer 31

yeah but how do we do that

Answer 32

First, we now the curve is near y=1 when x is big.
Let's figure out where it is when x is a large negative number

if we pick x= -100, and plug that into the equation what do we get for y ?

Answer 33

\[ y = \frac{-102 \cdot -103}{-101 \cdot -99} \]
it will be about +1.05 (above the line y=1)

Answer 34

ok. what does that mean?

Answer 35

I am sorry I am being difficult I have a massive headache and cant be done with school till this is done

Answer 36

as you go more and more negative the curve gets closer to 1 (but never reaches it...