Parabola help

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December 7th 2008, 03:43 PM
NeedHelp18

Parabola help

Consider the parabola defined by the equation y = -x^2 + 6.

Find: axis of symmetry: i got x = 0

Find: vertex: i got (0,6)

Express the distance as a function in x between the origin and point P=(x,y) which is on the parabola on the first quadrant: okay i have no idea how to do this one.

thanks for any help:)
December 7th 2008, 03:47 PM
mr fantastic

Quote:

Originally Posted by NeedHelp18

Consider the parabola defined by the equation y = -x^2 + 6.

Find: axis of symmetry: i got x = 0

Find: vertex: i got (0,6)

Express the distance as a function in x between the origin and point P=(x,y) which is on the parabola on the first quadrant: okay i have no idea how to do this one.

thanks for any help:)

A general point on the parabola has coordinates (x, y), that is, (x, -x^2 + 6). The origin has coordinate (0, 0). Now apply the usual formula for the distance between two points.
December 7th 2008, 04:09 PM
NeedHelp18

f(x) = √(x²+x^4 + 36) for 0≤x≤√6

is this the answer?
December 7th 2008, 04:27 PM
Skalkaz

Dear NeedHelp18,

I hope you don't belive so: $(a-b)^2 = a^2 + b^2$

It would be pity!
December 7th 2008, 04:44 PM
NeedHelp18

wats the answer then? i still keep gettin the same thing.
December 7th 2008, 04:47 PM
Skalkaz

the rigth formula:

$(a-b)^2 = a^2 - 2*a*b + b^2$
December 7th 2008, 06:49 PM
mr fantastic

Quote:

Originally Posted by NeedHelp18

f(x) = √(x²+x^4 + 36) for 0≤x≤√6

is this the answer?

No. $d = \sqrt{x^2 + (- x^2 + 6)^2} = \, ....$