distance as a function of x between (3,0) and (x,x^2) on parabola

[confused_07]

Express he distance between the point (3,0) and the point P(x,y) of the parabola y=x^2 as a function of x.

The function for the parabola is f(x) = x^2, that I know.
Would you use the difference formula: d= sqrt[(x_2-x_1)^2 - (y_2-y-1)^2] ?

i.e.: d= sqrt[(x-3)^2 - (y-0)^2]

 

[soroban]

Re: distance as a function of x

Hello, confused_07!

Express he distance between the point (3,0)
and the point P(x,y) of the parabola \(\displaystyle y\:=\:x^2\) as a function of \(\displaystyle x\).

The function for the parabola is: \(\displaystyle f(x) \,=\, x^2\), that I know.

Would you use the distance formula: \(\displaystyle \:d\:=\:\sqrt{(x_2-x_1)^2\,+\,(y_2-y_1)^2}\) ? . Yes!

i.e.: \(\displaystyle \:d\:=\:\sqrt{(x-3)^2\,+\,(y-0)^2}\;\) Right!

Just replace \(\displaystyle y\) with \(\displaystyle x^2\) . . .

\(\displaystyle d \;=\;\sqrt{(x\,-\,3)^2\,+\,(x^2\,-\,0)^2} \;=\;\sqrt{(x\,-\,3)^2\,+\,x^4}\)

 

[teejay207]

Hello, confused_07!


Just replace \(\displaystyle y\) with \(\displaystyle x^2\) . . .

\(\displaystyle d \;=\;\sqrt{(x\,-\,3)^2\,+\,(x^2\,-\,0)^2} \;=\;\sqrt{(x\,-\,3)^2\,+\,x^4}\)

\(\displaystyle d \;=\;\sqrt{(x\,-\,3)^2\,+\,(x^2\,-\,0)^2} \;=\;\sqrt{(x\,-\,3)^2\,+\,x^4}\)