Question

Find the following for the two functions:
-domain and range
-x and y intercepts
-horizontal asymptote(s)
-vertical asymptote(s)

1. 2/x^2-2x-3

2. x^2+x-2/x^2-3x-4

Answer 1

So far I think I am correct, but idk
1. domain: 2 +/- sqrt3/2
range: all real numbers
x intercept: none
y intercept: -.67
horizontal: 0
Vertical: -1

Answer 2

\[\frac{2}{(x-3)(x+1)}\] domain is all numbers except \(x=3,x=-1\)

Answer 3

no \(x\) intercept because it is never zero (since the numerator is 2)

Answer 4

oh okay and I think that there is another vertical intercept, but I cannot seem to find it

Answer 5

if \(x=0\) then \(y=-\frac{2}{3}\) for the \(y\) intercept

Answer 6

vertical asymptotes at the zeros of the denominator, namely \(x=-1,x=3\)

Answer 7

horizontal asymptote what you said, \(y=0\)

Answer 8

alright for the secong equation is it
domain and range is all real numbers

Answer 9

range is not all real numbers however

Answer 10

for example , this can never be 0, so zero is not in the range
i believe the range is \((-\infty,-\frac{1}{2})\cup (0,\infty)\) but that takes some checking

Answer 11

easy with calc, somewhat harder with algebra (i am still talking about the first one)

Answer 12

\[y=\frac{2}{x^2-2x+3}\]
\[y(x^2-2x-3)=2\]
\[yx^2-2yx-(3y+2)=0\] solve the quadratic for \(x\)

Answer 13

okay for the equation 2
\[\frac{ x^2+x-2 }{ x^2-3x-4 }\]
would the domain be all real numbers?

Answer 14

no, the denominator cannot be zero

Answer 15

\(x^2-3x-4=0\) or \((x-4)(x+1)=0\) so domain does not contain \(4\) or \(-1\)

Answer 16

oh okay is y intercept: 1/2
x intercept: -1.5 and 3.5

Answer 17

it is probably easier if you write this in factored form as
\[\frac{(x+2)(x-1)}{(x-4)(x+1)}\] then you can find what you need

Answer 18

two \(x\) intercepts, at \(x=-2,x=1\)

Answer 19

Horixontal asymptote is 1
and vertical is -1 and 0
right?