# Parabola help

• Dec 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:)
• Dec 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.
• Dec 7th 2008, 04:09 PM
NeedHelp18
f(x) = √(x²+x^4 + 36) for 0≤x≤√6

• Dec 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!
• Dec 7th 2008, 04:44 PM
NeedHelp18
wats the answer then? i still keep gettin the same thing.
• Dec 7th 2008, 04:47 PM
Skalkaz
the rigth formula:

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

Originally Posted by NeedHelp18
f(x) = √(x²+x^4 + 36) for 0≤x≤√6

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