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

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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

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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
(a) x^2-6x-y=0 we can solve this for x by using the quadratic formula \[x=\frac{6 \pm \sqrt{36-4(-y)}}{2}=\frac{6 \pm \sqrt{36+4y}}{2}\]
what does y have to be in order for x to exist?

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I am not sure? this question is confusing
can you have a negative under the square root?
no
you want 36+4y to be bigger than or equal to zero, correct?
yes correct
so solve this for y: 36+4y>=0 and you will have your range
-9 >= y
how do i write the range
4y>=-36 so y>=-9
what is the domain?
solve for y
to find range we solve for x to find domain we solve for y so solve the orignal equation for y
so y >= -9 is the range or domain?
are you still here?
i think you are supposed to solve each of these for y and think of y as a function of x
but there is something wrong here because you have these each set = 0
that is what my project say
y is range we got y>=-9
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
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
now*
not no lol
LOL u guys are great help
second one. \[4x^2-xy+1=0\] \[xy=4x^2+1\] \[y=\frac{4x^2+1}{x}\] domain all real numbers except 0
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
so the second one does not have a range
range is also all real numbers except 0 because the numerator is \[4x^2+1\] which cannot be 0
no, range is just the possible y values. y can be anything except 0 since the numerator of your fraction is never 0
range is [-4,4] you find once you solve for x
oops really? i will try it.
oops o did somehting wrong its (-inf,-4]U[4,inf)
is this for the second problem
the problem that we are now we are finding range
how do you like that. you get \[y^2-16\]under the radical so \[y^2-16\geq 0\] who knew?
lol satellitle
you remember all that inverse crap right?
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!
so the range is (-inf, -4)U(4,inf)
is the last one \[5x^2y-x+2=0\]?
yes you have the range correct.
ok
if i have the last one correct is is \[y=\frac{x-2}{5x^2}\] domain all numbers except 0
and the range?
honestly i am little confused by this. myininaya said solve for x. i can try it.
oh ok!
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.
so i think i will stick to that answer, and go to bed.
you want me to check your range? lol which one is this?
ok the last one
last one. i have confused myself.
if i didnt make a mistake here is the range
1 Attachment
i must be tired if i cannot multiply 4 by 5 by 2! i best retire.
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
ok maybe but i am still skeptical. what do you get if x = 2?
hmm you get (2,0)
looks like if x = 2, then y = 0 original equation is \[5x^2y-x+2=0\] replace x by 2 get \[20y-2+2=0\] \[20y=0\] \[y=0\]
i should switch back to windows and run maple, graph the thing and then i would be more convinced.
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
so the range is (-inf,1/40]
i am afraid i do not believe that either.
i believe in it lol
as x goes to infinity \[\frac{x-2}{5x^2}\] goes to zero
right there is a horizontal asymptote
let x = 1000000
bet you get something closer to 0 than 1/40
oh wait. maybe it is always negative. oops. ok possible
ok i am beginning to believe it. may be always less than 1/40. tomorrow i graph !
hey satelittle you know what would really convince you? i do
maple
you know how to find the maximum right? that derivative crap?
ok here we go.
so you do :) yay!
f'(x)=[5x^2-10x(x-2)]/[(5x^2)]^2=[-5x^2+20x]/(25x^4) a critical numbers isfour f(4)=1/40
well it is nice and late and i have had a few beers. but it looks like extreme value at 0 and 4
well 0 doesn't exist in the orignal function
0 is out, 4 gives.... holy carp, what do you know.
lol
learn something new every day. thanks.
who needs maple
me. use it all the time when i am not running linux
well it is a nice friend to have
and so to bed, before i make more of a fool of myself. take care
later
i'm going too gn

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