Let P = (x, y) be a point on the graph of y = x^2 + 2.

Express the distance d from P to the point (2, 0) as a function of x.

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- Jan 19th 2007, 06:54 PMsymmetryDistance d from P
Let P = (x, y) be a point on the graph of y = x^2 + 2.

Express the distance d from P to the point (2, 0) as a function of x. - Jan 19th 2007, 07:52 PMticbol
Distance between two points (x1y1) and (x2,y2):

d = sqrt[(x2 -x1)^2 +(y2 -y1)^2] ---------------------------------(i)

or

d = sqrt[(difference of the x's)^2 +(difference of the y's)^2] ------(ii)

Meaning, (x1 -x2)^2 is also (x2 -x1)^2; (y2 -y1)^2 is same as (y1 -y2)^2.

The point P(x,y) on the graph of y = x^2 +2

is P(x,(x^2 +2)) in terms of x.

The other point is (2,0)

So,

d = sqrt[(x -2)^2 +((x^2 +2) -0)

d = sqrt[(x-2)^2 +(x^2 +2)]

d = sqrt[(x^2 -4x +4) +(x^2 +2)]

d = sqrt[x^2 -4x +4 +x^2 +2]

d = sqrt[2x^2 -4x +6] --------------------answer.

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Edit.

Oooppps, wrong. I forgot to square the difference of the y's.

It should have been like this:

d = sqrt[(x -2)^2 +((x^2 +2) -0)^2]

d = sqrt[(x-2)^2 +(x^2 +2)^2]

d = sqrt[(x^2 -4x +4) +(x^4 +4x^2 +4)]

d = sqrt[x^2 -4x +4 +x^4 +4x^2 +4]

d = sqrt[x^4 +5x^2 -4x +8] --------------------final answer. - Jan 20th 2007, 05:24 AMsymmetryok
I thank you for every for the insight and math tips.

Basically, we apply the distance formula.

Good job!

Thank!