anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • 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
  • 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
  • myininaya
what does y have to be in order for x to exist?

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anonymous
  • anonymous
I am not sure? this question is confusing
myininaya
  • myininaya
can you have a negative under the square root?
anonymous
  • anonymous
no
myininaya
  • myininaya
you want 36+4y to be bigger than or equal to zero, correct?
anonymous
  • anonymous
yes correct
myininaya
  • myininaya
so solve this for y: 36+4y>=0 and you will have your range
anonymous
  • anonymous
-9 >= y
anonymous
  • anonymous
how do i write the range
myininaya
  • myininaya
4y>=-36 so y>=-9
anonymous
  • anonymous
what is the domain?
myininaya
  • myininaya
solve for y
myininaya
  • myininaya
to find range we solve for x to find domain we solve for y so solve the orignal equation for y
anonymous
  • anonymous
so y >= -9 is the range or domain?
anonymous
  • anonymous
are you still here?
anonymous
  • anonymous
i think you are supposed to solve each of these for y and think of y as a function of x
anonymous
  • anonymous
but there is something wrong here because you have these each set = 0
anonymous
  • anonymous
that is what my project say
myininaya
  • myininaya
y is range we got y>=-9
anonymous
  • 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
  • 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
  • myininaya
now*
myininaya
  • myininaya
not no lol
anonymous
  • anonymous
LOL u guys are great help
anonymous
  • 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
  • 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
  • anonymous
so the second one does not have a range
anonymous
  • anonymous
range is also all real numbers except 0 because the numerator is \[4x^2+1\] which cannot be 0
anonymous
  • 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
  • myininaya
range is [-4,4] you find once you solve for x
anonymous
  • anonymous
oops really? i will try it.
myininaya
  • myininaya
oops o did somehting wrong its (-inf,-4]U[4,inf)
anonymous
  • anonymous
is this for the second problem
myininaya
  • myininaya
the problem that we are now we are finding range
anonymous
  • anonymous
how do you like that. you get \[y^2-16\]under the radical so \[y^2-16\geq 0\] who knew?
myininaya
  • myininaya
lol satellitle
myininaya
  • myininaya
you remember all that inverse crap right?
anonymous
  • 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
  • anonymous
so the range is (-inf, -4)U(4,inf)
anonymous
  • anonymous
is the last one \[5x^2y-x+2=0\]?
anonymous
  • anonymous
yes you have the range correct.
anonymous
  • anonymous
ok
anonymous
  • anonymous
if i have the last one correct is is \[y=\frac{x-2}{5x^2}\] domain all numbers except 0
anonymous
  • anonymous
and the range?
anonymous
  • anonymous
honestly i am little confused by this. myininaya said solve for x. i can try it.
anonymous
  • anonymous
oh ok!
anonymous
  • 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
  • anonymous
so i think i will stick to that answer, and go to bed.
myininaya
  • myininaya
you want me to check your range? lol which one is this?
myininaya
  • myininaya
ok the last one
anonymous
  • anonymous
last one. i have confused myself.
myininaya
  • myininaya
if i didnt make a mistake here is the range
1 Attachment
anonymous
  • anonymous
i must be tired if i cannot multiply 4 by 5 by 2! i best retire.
myininaya
  • 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
  • anonymous
ok maybe but i am still skeptical. what do you get if x = 2?
myininaya
  • myininaya
hmm you get (2,0)
anonymous
  • 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
  • anonymous
i should switch back to windows and run maple, graph the thing and then i would be more convinced.
myininaya
  • 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
  • myininaya
so the range is (-inf,1/40]
anonymous
  • anonymous
i am afraid i do not believe that either.
myininaya
  • myininaya
i believe in it lol
anonymous
  • anonymous
as x goes to infinity \[\frac{x-2}{5x^2}\] goes to zero
myininaya
  • myininaya
right there is a horizontal asymptote
anonymous
  • anonymous
let x = 1000000
anonymous
  • anonymous
bet you get something closer to 0 than 1/40
anonymous
  • anonymous
oh wait. maybe it is always negative. oops. ok possible
anonymous
  • anonymous
ok i am beginning to believe it. may be always less than 1/40. tomorrow i graph !
myininaya
  • myininaya
hey satelittle you know what would really convince you? i do
anonymous
  • anonymous
maple
myininaya
  • myininaya
you know how to find the maximum right? that derivative crap?
anonymous
  • anonymous
ok here we go.
myininaya
  • myininaya
so you do :) yay!
myininaya
  • 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
  • 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
  • myininaya
well 0 doesn't exist in the orignal function
anonymous
  • anonymous
0 is out, 4 gives.... holy carp, what do you know.
myininaya
  • myininaya
lol
anonymous
  • anonymous
learn something new every day. thanks.
myininaya
  • myininaya
who needs maple
anonymous
  • anonymous
me. use it all the time when i am not running linux
myininaya
  • myininaya
well it is a nice friend to have
anonymous
  • anonymous
and so to bed, before i make more of a fool of myself. take care
myininaya
  • myininaya
later
myininaya
  • myininaya
i'm going too gn

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