intercepts

orchidpsycho

New member
Joined
Sep 26, 2005
Messages
41
was asked to make a table of values and sketch graph for the equation. find x and y intercepts and test for symmetry

y=x^2-3

got the graph with points (0,3) (1,-2) (2,1) (-1,-2) (-2,1) (-3,6) (3,6)

its symmetrical with respect to the y axis because the equation is unchanged when x is replaced with -x

the problem I am wondering is when i find the intercepts I get

y intercept= -3 (0,-3)

but the x I get +- sqrt[3] is that right. wasnt sure how I would present that?

thanks for any help.
 
Hello, orchidpsycho!

Your work absolutely correct.
. . And you noted the symmetry . . . nice touch!

but the x I get +- sqrt[3] is that right. wasnt sure how I would present that?
It's not our fault that the intercepts are irrational numbers . . .
. . . . . . . . . . . . . . . _
We can write: . (±√3, 0)

If you're making a sketch, you can approximate it: .(±1.7, 0) is close enough.
 
thanks.

Thought it was right but up until now I have'nt had any irrational numbers come into play.

Thanks for all the help. I don't know what I would do without people like you to help me along.


One more question. How did you get that square root symbol? Not sure how to do that?
 
tutorial

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Hello, orchidpsycho!

How did you get that square root symbol?
There are codes available; I have a partial list.

You precede them with &# (ampersand, 'number') and follow them with ; (semicolon)

√ 8730

θ 952

≠ 8800

± 177

° 176 (degree)

→ 8594
 
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