Question

use algebraic methods to determine the domain and range for the function from to defined by the given equation.

a. x^2 – 6x –y =0
b. 4x^2 – xy +1 = 0
c. 5x^2(y) – x + 2 = 0

Answer 1

use algebraic methods to determine the domain and range for the function from x to y defined by the given equation.

a. x^2 – 6x –y =0
b. 4x^2 – xy +1 = 0
c. 5x^2(y) – x + 2 = 0

Answer 2

(a) x^2-6x-y=0
we can solve this for x by using the quadratic formula
\[x=\frac{6 \pm \sqrt{36-4(-y)}}{2}=\frac{6 \pm \sqrt{36+4y}}{2}\]

Answer 3

what does y have to be in order for x to exist?

Answer 4

I am not sure? this question is confusing

Answer 5

can you have a negative under the square root?

Answer 6

no

Answer 7

you want 36+4y to be bigger than or equal to zero, correct?

Answer 8

yes correct

Answer 9

so solve this for y: 36+4y>=0 and you will have your range

Answer 10

-9 >= y

Answer 11

how do i write the range

Answer 12

4y>=-36
so y>=-9

Answer 13

what is the domain?

Answer 14

solve for y

Answer 15

to find range we solve for x
to find domain we solve for y
so solve the orignal equation for y

Answer 16

so y >= -9 is the range or domain?

Answer 17

are you still here?

Answer 18

i think you are supposed to solve each of these for y
and think of y as a function of x

Answer 19

but there is something wrong here because you have these each set = 0

Answer 20

that is what my project say

Answer 21

y is range we got y>=-9

Answer 22

ok fine. so first one is \[x^2-6x-y=0\]
so \[y=x^2-6x\]
as a function
\[f(x)=x^2-6x\]
domain all real numbers since it is a polynomial
range is -9 to infinity

Answer 23

no we find domain solve x^2-6x-y=0 for x
and you will see the domain is all real numbers because x^2-6x is a polynomial

Answer 24

now*

Answer 25

not no lol

Answer 26

LOL u guys are great help

Answer 27

second one.
\[4x^2-xy+1=0\]
\[xy=4x^2+1\]
\[y=\frac{4x^2+1}{x}\]
domain all real numbers except 0

Answer 28

all you have to do is solve both for x and y
when you solve for x you can find the range
when you solve for y you can find the domain

Answer 29

so the second one does not have a range

Answer 30

range is also all real numbers except 0 because the numerator is \[4x^2+1\] which cannot be 0

Answer 31

no, range is just the possible y values. y can be anything except 0 since the numerator of your fraction is never 0

Answer 32

range is [-4,4] you find once you solve for x

Answer 33

oops really? i will try it.

Answer 34

oops o did somehting wrong its (-inf,-4]U[4,inf)

Answer 35

is this for the second problem

Answer 36

the problem that we are now we are finding range

Answer 37

how do you like that. you get \[y^2-16\]under the radical so \[y^2-16\geq 0\]
who knew?

Answer 38

lol satellitle

Answer 39

you remember all that inverse crap right?

Answer 40

oh of course. this thing is a rational function with a slant asymptote. so it does skip values, never between -4 and 4. must be late!

Answer 41

so the range is (-inf, -4)U(4,inf)

Answer 42

is the last one \[5x^2y-x+2=0\]?

Answer 43

yes you have the range correct.

Answer 44

ok

Answer 45

if i have the last one correct is is
\[y=\frac{x-2}{5x^2}\]
domain all numbers except 0

Answer 46

and the range?

Answer 47

honestly i am little confused by this. myininaya said solve for x. i can try it.

Answer 48

oh ok!

Answer 49

if i use the quadratic formula i see under the radical
\[1-20y\]
meaning \[1-20y\geq 0\]
\[y\leq \frac{1}{20}\]
would held if i had the picture of this thing to see if that makes sense.

Answer 50

so i think i will stick to that answer, and go to bed.

Answer 51

you want me to check your range? lol
which one is this?

Answer 52

ok the last one

Answer 53

last one. i have confused myself.

Answer 54

if i didnt make a mistake here is the range

Answer 55

i must be tired if i cannot multiply 4 by 5 by 2!
i best retire.

Answer 56

yeah you were close then we also have y in the denominator
and we know we cannot divide by 0
and when y is 0 the denominator is 0 so our range does not include y=0

Answer 57

ok maybe but i am still skeptical. what do you get if x = 2?

Answer 58

hmm you get (2,0)

Answer 59

looks like if x = 2, then y = 0
original equation is
\[5x^2y-x+2=0\]
replace x by 2 get
\[20y-2+2=0\]
\[20y=0\]
\[y=0\]

Answer 60

i should switch back to windows and run maple, graph the thing and then i would be more convinced.

Answer 61

well and also the quadratic formula just failes when a=0
in this case when 5yx^2=0 since y=0 a=0
so the quadratic formula won't work for that part of the answer

Answer 62

so the range is (-inf,1/40]

Answer 63

i am afraid i do not believe that either.

Answer 64

i believe in it lol

Answer 65

as x goes to infinity
\[\frac{x-2}{5x^2}\] goes to zero

Answer 66

right there is a horizontal asymptote

Answer 67

let x = 1000000

Answer 68

bet you get something closer to 0 than 1/40

Answer 69

oh wait. maybe it is always negative. oops. ok possible

Answer 70

ok i am beginning to believe it. may be always less than 1/40. tomorrow i graph
!

Answer 71

hey satelittle you know what would really convince you?
i do

Answer 72

maple

Answer 73

you know how to find the maximum right?
that derivative crap?

Answer 74

ok here we go.

Answer 75

so you do :) yay!

Answer 76

f'(x)=[5x^2-10x(x-2)]/[(5x^2)]^2=[-5x^2+20x]/(25x^4)
a critical numbers isfour
f(4)=1/40

Answer 77

well it is nice and late and i have had a few beers. but it looks like extreme value at 0 and 4

Answer 78

well 0 doesn't exist in the orignal function

Answer 79

0 is out, 4 gives....
holy carp, what do you know.

Answer 80

lol

Answer 81

learn something new every day. thanks.

Answer 82

who needs maple

Answer 83

me. use it all the time when i am not running linux

Answer 84

well it is a nice friend to have

Answer 85

and so to bed, before i make more of a fool of myself. take care

Answer 86

later

Answer 87

i'm going too gn