2007-01-30 13:55:46 · 8 answers · asked by Ashley 2 in Science & Mathematics ➔ Mathematics
No, it is not.
If you write it in "y equals" form, you get (I'll go through the steps)
y^2 = x^2 - 1
y = +/- sqrt(x^2 - 1)
Now, y = sqrt(x^2-1) is a function, and so is y = -sqrt(x^2 -1).
But the original equation you asked about is both of those. And if you graph both of the parts of the equation, you end up with a hyperbola.
This ISN'T a function because if you draw a vertical line, you will intersect the curve twice, once in the top half, and once in the bottom half. (That's one of the tests for something being a function.)
2007-01-30 14:06:32 · answer #1 · answered by branzillie 2 · 0⤊ 0⤋
no.
y=sqrt(x^2-1)
In order to be a function, one of the variables has to be uniquely dependent on the other variable. Calling "square root" a function is a misnomer because "square root" returns the plus and minus results. This is an artifact of computer programming languages which typically return the absolute value of the square root.
The formula is for a hyperbola.
2007-01-30 14:00:39 · answer #2 · answered by J C 5 · 0⤊ 0⤋
No its not a function. If you were to make it equal to y it would have to be + or - y. y^2=x^2-1 Then square root both sides and the answers to square roots are + or - the number.
2007-01-30 14:00:29 · answer #3 · answered by Anonymous · 0⤊ 0⤋
it is a function of two variables. see below
f(x,y)=x^2-y^2-1
2007-01-30 14:05:42 · answer #4 · answered by A J 1 · 0⤊ 0⤋
Yeah. It's a function with two unknowns.
2007-01-30 14:00:49 · answer #5 · answered by cloud_strife 2 · 0⤊ 1⤋
yes
2007-01-30 13:58:39 · answer #6 · answered by Marie 2 · 0⤊ 2⤋
i think so
i get a V when i graph it but i thought it wuz supposed to graph as a circle....
2007-01-30 14:03:06 · answer #7 · answered by adr k 2 · 0⤊ 1⤋
yes it is
2007-01-30 14:00:05 · answer #8 · answered by Anonymous · 0⤊ 1⤋