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

- anonymous

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- anonymous

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

- myininaya

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

- myininaya

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

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## More answers

- anonymous

I am not sure? this question is confusing

- myininaya

can you have a negative under the square root?

- anonymous

no

- myininaya

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

- anonymous

yes correct

- myininaya

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

- anonymous

-9 >= y

- anonymous

how do i write the range

- myininaya

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

- anonymous

what is the domain?

- myininaya

solve for y

- myininaya

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

- anonymous

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

- anonymous

are you still here?

- anonymous

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

- anonymous

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

- anonymous

that is what my project say

- myininaya

y is range we got y>=-9

- anonymous

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

- myininaya

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

- myininaya

now*

- myininaya

not no lol

- anonymous

LOL u guys are great help

- anonymous

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

- myininaya

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

- anonymous

so the second one does not have a range

- anonymous

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

- anonymous

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

- myininaya

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

- anonymous

oops really? i will try it.

- myininaya

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

- anonymous

is this for the second problem

- myininaya

the problem that we are now we are finding range

- anonymous

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

- myininaya

lol satellitle

- myininaya

you remember all that inverse crap right?

- anonymous

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!

- anonymous

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

- anonymous

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

- anonymous

yes you have the range correct.

- anonymous

ok

- anonymous

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

- anonymous

and the range?

- anonymous

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

- anonymous

oh ok!

- anonymous

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.

- anonymous

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

- myininaya

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

- myininaya

ok the last one

- anonymous

last one. i have confused myself.

- myininaya

if i didnt make a mistake here is the range

##### 1 Attachment

- anonymous

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

- myininaya

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

- anonymous

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

- myininaya

hmm you get (2,0)

- anonymous

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

- anonymous

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

- myininaya

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

- myininaya

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

- anonymous

i am afraid i do not believe that either.

- myininaya

i believe in it lol

- anonymous

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

- myininaya

right there is a horizontal asymptote

- anonymous

let x = 1000000

- anonymous

bet you get something closer to 0 than 1/40

- anonymous

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

- anonymous

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

- myininaya

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

- anonymous

maple

- myininaya

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

- anonymous

ok here we go.

- myininaya

so you do :) yay!

- myininaya

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

- anonymous

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

- myininaya

well 0 doesn't exist in the orignal function

- anonymous

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

- myininaya

lol

- anonymous

learn something new every day. thanks.

- myininaya

who needs maple

- anonymous

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

- myininaya

well it is a nice friend to have

- anonymous

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

- myininaya

later

- myininaya

i'm going too gn

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