Question

Need some intuitive understanding or some graphs...

Answer 1

First i wanna know is that why’s \[Y = \sqrt{-x^2+4}\] so similar to \[X^2+Y^2 = 4\]
Second, how did a simple funcnction graph of \[Y = \sqrt{x} \] or \[Y = \sqrt{-x} \] turned into circular graph like \[Y = \sqrt{-x^2 +4}\] why’s that so similar to \[X^2 + y^2 = 4\]

Answer 2

Also why’s the graph of \[Y = \sqrt{x^2}\]so distinctively different from the rest?

Answer 3

Why’s ^^^^^^ so similar to absolute function graph?

Answer 4

the first one

Y=sqrt(−x2+4) square both sides and get y^2 = -x^2 +4 yes ?
so similar to
X2+Y2=4

Answer 5

and in the second step assume to both sides +x^2

and will get x^2 +y^2 = 4

do you understand ?

Answer 6

Not really... don’t worry about it... i’m typing out the stuff for my teachers to look at. I prefer them to help me understand :) hope u dont mind :)

Answer 7

Also i need to know why’s the behavoir of root function graphs are so unpredictable...in a way. It’s not just a matter of simple *transformation of graph(s)* type scenario... like why’s the graph of \[Y = \sqrt{-x} \] reflected on to the second quadrant? Shouldn’t it be reflected down in the 4th quadrant like \[Y = x^2, y = -x^2\] do

Answer 8

What you're asking about is covered in transformations of functions. And if what you graph is coming up blank, it means you need to either zoom in or out.

Answer 9

But you won't be able to graph anything on that old eighties Casio

Answer 10

So for now, use Geogebra

Answer 11

Bro, i’m Literally on Desmos and i get nothing when i try to graph that function... i’ve Even zoomed in and out...

Answer 12

It doesn't graph because you're trying to graph the square root of a negative number.

Answer 13

The square root of a negative number is imaginary

Answer 14

The thing is that ...i was writing notes on how the original graph of \[Y = \sqrt{-x^2 +4}\] is transformed from its parent function and by the time i got to \[\sqrt{-x^2}\] part, errthing blanked out on me..

Answer 15

\(y = \sqrt{-x^2}\) is always imaginary, so it graphs no points, but \(y = \sqrt{-x^2+4}\) will graph a few points. Enough to create a semi-circle.

Answer 16

And why’s that so similar to x^2 +y^2 = 4

Answer 17

Because \(x^2 + y^2 = 4\) is the graph of a circle

Answer 18

Because when i look at \[Y = \sqrt{-x+4}\] i take that as a graph that is shifted 4 units to the left...but it doesn’t ;-;

Answer 19

If you have \(x^2 + y^2 = 4\) then subtract \(x^2\) from both sides and square root both sides you get \(y = \sqrt{4 - x^2}\) which is the top half of the circle. \(y = -\sqrt{4 - x^2}\) is the bottom half of the circle.

Answer 20

Wait let me draw that... one sec

Answer 21

You should review Transformations of Functions in IXL

Answer 22

I have function transformations coming up on page 4 i believe..

Answer 23

To learn the difference between \(y = -\sqrt{x}\) and \(y = \sqrt{-x}\)

Answer 24

Those are two different transformations of \(y = \sqrt{x}\)

Answer 25

One reflects \(\sqrt{x}\) over the vertical axis, the other reflects \(\sqrt{x}\) over the horizontal axis.

Answer 26

One should reflect the parent function in the 2nd quadrant and the other flips it in the 4th?

Answer 27

Okie dokie...

Answer 28

Soo how do i figure out the domain and range of function \[Y= \sqrt{-x^2+4}\]

Answer 29

Well, one way to figure it out is by graphing. The other way is by attempting to graph points. We know that the square root of a negative number is imaginary so the domain must be \(-4 \le x \le 4\).

Answer 30

Plz make it easy

Answer 31

Take the values of the domain and plug them in and see what you get for \(y\) values.

Answer 32

From there, you can determine the range.

Answer 33

Way way way way way.. i need to show you something...let me draw.

Answer 34

Did you install geogebra yet?

Answer 35

Sorry the domain is \(-2 \le x \le 2\) my bad

Answer 36

I have symbolab

Answer 37

Yuhh, the domain should be between -2 and 2

Answer 38

I know about symbolab. That's good but geogebra is the best of the best of graphing utilities.

Answer 39

Hopefully it’s free. Or else i might haff to getta pirated copy

Answer 40

It's definitely free.

Answer 41

Swee :)

Answer 42

Wait, let me finish off this question...brb And THANKS :)

Answer 43

Thanks everyone:)