domain and range...yes (1 Viewer)

[crammy90]

graph y = root(4-x^2) and state range.
heres how i have taught myself to do it from answers
i) y = o, solve for x to get x=+-2
x = 0 to solve and get y=+-2
so this is a circle right/
but then they take only the top half...is this because of the root?
are we meant to know a semi circle is a root over an equation or seomthing?
and how do we know if the range is outside the half circle or inside
thanks heaps lol

 

[stormz89]

its a semi circle can't remember how but i do remember if its in a square root form then its a semi circle. and the range is always inside the circle so its 2.

 

[Just.Snaz]

well for y = root(4 - x^2), since you have a root.. whatever's under the root MUST be > or = 0
if you solve 4 - x^2 >= 0 you get -2 <= x => 2
now, the root of anything always has to be positive, correct? and under the root COULD = 0 so y > or = 0.. so yes it is a circle but your range is y >=0 so it is only the top half of the circle, ie a semi circle.

as a general rule, something in that form is always a semi circle..

 

[stormz89]

See I understand this but as a second thought, isn't the square root of a number such as 4 is +-2.

I know that a surd has to be positive to have a root. I also know that your are correct and its the general form of a semi circle.


Is this a question from past HSC? I've seen it before somewhere.

 

[Imagination104]

well for y = root(4 - x^2), since you have a root.. whatever's under the root MUST be > or = 0
if you solve 4 - x^2 >= 0 you get -2 <= x => 2
now, the root of anything always has to be positive, correct? and under the root COULD = 0 so y > or = 0.. so yes it is a circle but your range is y >=0 so it is only the top half of the circle, ie a semi circle.

as a general rule, something in that form is always a semi circle..

Isn't the range 0=< y =<2? Since the radius of the semi-circle is 2.

 

[alez]

yeah its only 0 to 2 for y
as root(4-x) become root 4, and its only pos
only between -2 and 2 when its y^2, not when the root is there

 

[Just.Snaz]

Isn't the range 0=< y =<2? Since the radius of the semi-circle is 2.

yes that is true. my mistake. I was trying to point out that it becomes a semi circle because you have to neglect the bottom half since y >= 0.

See I understand this but as a second thought, isn't the square root of a number such as 4 is +-2.

I know that a surd has to be positive to have a root. I also know that your are correct and its the general form of a semi circle.


Is this a question from past HSC? I've seen it before somewhere.

when you have x^2 = 4, then x = +-2 because if you sub either value, you get 4
but if you have x =root4, x must be 2

 

[crammy90]

when you have x^2 = 4, then x = +-2 because if you sub either value, you get 4
but if you have x =root4, x must be 2

woah
but u can make x^2 = 4 into x =root4...
so confused aye lol

 

[crammy90]

heres another
state domain and range
y=2root(25-x^2)
what different does the 2 make?

 

[eskimoh]

i think the range becomes between 0 and 10 but the domain remains betweeen plusminus 5

 

[Just.Snaz]

woah
but u can make x^2 = 4 into x =root4...
so confused aye lol

no x^2 = 4
x = +- root4
x = +- 2
because as I said before, you sub in either x =2 or x = -2 into the original equation and you'll still get 4
there is however a bit of a contradiction but but it only creates more confusion and it doesn't matter.. just learn it and get the marks
so simply, x = root 4 = 2
but x^2 = 4, x = +-2

heres another
state domain and range
y=2root(25-x^2)
what different does the 2 make?

as someone said,
the domain would be done the same way.. (whatever is under the square root is done in the same way) and it'd end up being -5 <= x <= 5
and the range would be 0 <= y =< 10
you can get this by finding the y-intercept and then sketching it. And you get the y-intercept because you know that the centre is (0,0) since it's in the form of a semi circle and the domain is - 5 <= x =< 5 so the max y value would have x = 0.