Question

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

Answer 1

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.

Answer 2

have I made you understood??

Answer 3

where you get y^2/9-x^2/9=1 from? is that equation for hyperbola?

Answer 4

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

Answer 5

oh i see now, i was thinking the x was in the denominator for y² because of computer writing ;)

Answer 6

haha no, ken is just showing you that the equation is actually a hyperbola, and he proved it by simplifying

Answer 7

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

Answer 8

the former has its foci on the x-axis, and the latter has its foci on the y-axis.

Answer 9

kk, all good now. cheers

Answer 10

cheers(^^)

Answer 11

www.aceyourcollegeclasses.com is a great place to post your questions