# I need this explained....

• Dec 8th 2010, 12:39 PM
I need this explained....
Hello,
My notes have y=sqrt x^2-16
Now it's -4 and 4 on the x axis, but just the top part of the y curve, as it starts with that. going out to each side.

So the fixing up the equation above is y^2=x^2-16
so y^2-x^2=-16 which turns into x^2-y^2=16.
Now x^2=16 is x=4, But for y it's y = -4, but how does that work? I need x -4,+4.

I don't understand how to come up with the -4 on x.
Does this make sense? it's hard to write out without a graph.

Graphing Calculator

If you go to this graphing online you will see what I am talking about keying in the above equation in bold

If someone can explain this would be great, thanks.
Joanne

just trying to understand it, because I am not getting it.
• Dec 8th 2010, 01:05 PM
Quote:

Originally Posted by bradycat
Hello,
My notes have y=sqrt x^2-16
Now it's -4 and 4 on the x axis, but just the top part of the y curve, as it starts with that. going out to each side.

So the fixing up the equation above is y^2=x^2-16
so y^2-x^2=-16 which turns into x^2-y^2=16.
Now x^2=16 is x=4, But for y it's y = -4, but how does that work? I need x -4,+4.

I don't understand how to come up with the -4 on x.
Does this make sense? it's hard to write out without a graph.

Graphing Calculator

If you go to this graphing online you will see what I am talking about keying in the above equation in bold

If someone can explain this would be great, thanks.
Joanne

just trying to understand it, because I am not getting it.

$y=\sqrt{x^2-16}$

means that "y" is the positive square root of $x^2-16$

so $\left(x^2-16\right)\ \ge\ 0$

$\Rightarrow\ x^2\ \ge\ 16\Rightarrow\ x\ \ge\ 4$ if x is positive

or $x\ \le\ -4$ if x is negative.

What exactly are you looking for ?
• Dec 8th 2010, 01:08 PM
masters
Quote:

Originally Posted by bradycat
Hello,
My notes have y=sqrt x^2-16
Now it's -4 and 4 on the x axis, but just the top part of the y curve, as it starts with that. going out to each side.

So the fixing up the equation above is y^2=x^2-16
so y^2-x^2=-16 which turns into x^2-y^2=16.
Now x^2=16 is x=4, But for y it's y = -4, but how does that work? I need x -4,+4.

I don't understand how to come up with the -4 on x.
Does this make sense? it's hard to write out without a graph.

Graphing Calculator

If you go to this graphing online you will see what I am talking about keying in the above equation in bold

If someone can explain this would be great, thanks.
Joanne

just trying to understand it, because I am not getting it.

Hi brandycat,

It seems to me that you are trying to find the x-intercepts of $y=\sqrt{x^2-16}$

Set y = 0 and solve for x.

$0=\sqrt{x^2-16}\:\:or \:\: \sqrt{x^2-16}=0$

Square both sides now.

$x^2-16=0$

$x^2=16$

$x=\pm 4$
• Dec 9th 2010, 01:55 AM
HallsofIvy
The graph of $x^2- y^2= 16$ is a hyperbola with two separated parts- one to the right of the y-axis, the other to the left.

The graph of $y= \sqrt{x^2- 16}$ is the upper half of that hyperbola and consists of two separate "wings". It meets the x-axis at x= 4 and x= -4 but does NOT cross the y- axis. In fact x is always either larger than or equal to 4 or less than or equal to -4.