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How do i draw a graph for this : y = sqrt (x2 -1) ? there seems to be no points in...

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sweetskittles24 | Student, Grade 11 | Salutatorian

Posted June 27, 2012 at 11:09 PM via web

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How do i draw a graph for this : y = sqrt (x2 -1) ? there seems to be no points in quadrant 2, why is this?

The thing is, for every negative x variable, you can square it and it will equal to a positive number (which will give you a point in quadrant 2). I have seen graphs for this equation, but there seems to be no points in quadrant 2, why is this?

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted June 28, 2012 at 4:39 AM (Answer #1)

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Graph `y=sqrt(x^2-1)` :

If `f(x)=sqrt(x^2-1)` , note that `f(-x)=f(x)` , so the graph is symmetric about the y-axis. In other words, the graph has a mirror image over the y-axis. So if you can graph the function in the 1st quadrant, the graph in the 2nd quadrant is the reflection across the y-axis.

The graph:

Note that the domain is `|x|>=1` and the range is `y>=0`

jerichorayel points out that this is a hyperbola; however this is incorrect since you had to square both sides -- this results in extraneous solutions, notable the half of the hyperbola in the lower quadrants.

Some graphing utilities may not accept negative inputs for x when taking the square root -- they would be unable to graph `y=sqrt(-x)` which is clearly the reflection of `y=sqrt(x)` across the x-axis. This is a limitation in their programming, not the way the graphs exist.

Sources:

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jerichorayel | College Teacher | (Level 1) Senior Educator

Posted June 28, 2012 at 3:57 AM (Answer #2)

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you can rewrite the equation as

x^2 - y^2 = 1  and this is a hyperbola.

 

x = +/- 1 so you plot the points at x = 1 and -1

y = 0

 

this is the plot from wolfram alpha.

x^2 - Y^2 = 1    ---> ste y = 0 to get the intercepts

x^2 - 0 = 1

x = +/- 1

do the same for y, but there would be no real solution

since it would become   sqrt(-1)

 

 

 

hope this helps.

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