Question

Algebra 2 help asap!?
1.find the horizontal o oblique asymptote of f(x)= -2x^2+3x+6/x+1
a. y=-1
b.y=-2
c.y=2x+3
d.y=-2x+5

2. what value is a discontinuity of x^2+5x+2/x^2+2x-35
a.x=-5
b.x=-7
c.x=-2
d.x=-1

3.what is the simplified form of 1/y=1/x/1/y+1/x?
a. y+x/y-x
b.y-x/y+x
c.x-y/x+y
d.x+y/x-y

Answer 1

@jim_thompson5910

Answer 2

@Directrix

Answer 3

@mathslover

Answer 4

lol

Answer 5

can you help me please! @misty1212

Answer 6

for the first one, divide

Answer 7

i have no idea how to do this @misty1212

Answer 8

divide
you get a quotient and a remainder
ignore the remainder, the quotient in the oblique asymptote
you can divide using polynomial long division (ick)
synthetic division (real easy)
or cheating (even easier)
http://www.wolframalpha.com/input/?i=%28+-2x^2%2B3x%2B6%29%2F%28x%2B1%29+

Answer 9

okay and for the second and third one?

Answer 10

actually you don't have to do any work
the leading coefficient is -2 so it has to be D for the first one

Answer 11

cool!

Answer 12

for the second one factor and cancel

Answer 13

or just factor the denominator lets check it

Answer 14

okay how do we do this

Answer 15

is it -7?

Answer 16

set
\[x^222x-35=0\] and solve

Answer 17

oops
\[x^2+2x-35=0\] solve for \(x\)

Answer 18

-7 and 5 but 5 is not an option so -7?

Answer 19

@misty1212

Answer 20

\[(x+7)(x-5)=0\\
x=-7,x=5\] yeah

Answer 21

thanks! can you help me with one more please!? @misty1212

Answer 22

Using a directrix of y=2 and a focus of (3,-4), what quadratic function is created?

Answer 23

sure what is the vertex?

Answer 24

it doesnt say

Answer 25

i know dear, you have to figure it out!! it is half way between the line \(y=2\) and the point \((3,-4)\)
what point is half way?

Answer 26

3,-1?

Answer 27

@misty1212