ƒ(x) = √(x²+9) Is this a semi circle? :S

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ƒ(x) = √(x²+9) Is this a semi circle? :S

Mathematics
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no it's not. you see y=√(x²+9)>0, also y^2/9-x^2/9=1, therefore the graph is half of the hyperbola whose focus is (0,3√2) and which is above the x-axis.
have I made you understood??
where you get y^2/9-x^2/9=1 from? is that equation for hyperbola?

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square y then move it over so that you have 9 = y^2-x^2 then divide by nine to get 1=y^2/9-x^2/9
oh i see now, i was thinking the x was in the denominator for y² because of computer writing ;)
haha no, ken is just showing you that the equation is actually a hyperbola, and he proved it by simplifying
yes. it;s like that. a standard hyperbola equation is x^2/a^2-y^2/b^2=1 or y^2/a^2-x^2/b^2=1
the former has its foci on the x-axis, and the latter has its foci on the y-axis.
kk, all good now. cheers
cheers(^^)
www.aceyourcollegeclasses.com is a great place to post your questions

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