## Trending News

Promoted

# How do I find the coordinates?

The problem is "Find the coordinates of the point on the graph of y=x^2 that is closest to the point (0,1) on the x-axis."

I think I need to change it to point-slope, and I know it involves a perpendicular slope, but I'm not sure how to make it through the entire thing so show work if possible. I've seen it done using the distance formula, but we haven't gone over it yet in class so I don't think I'm supposed to use it. Any help is appreciated.

### 1 Answer

Relevance

- NozarLv 68 years agoFavorite Answer
D^2=(1-x^2)^2 +x^2=1+x^4-2x^2 +x^2

x^4-x^2+1=D^2, the derivative is'

4x^3-2x=0

x=0,y=0,D^2=1,D=1

2x(2x^2-1)=0

2x^2=1, x^2=1/2, x=+&-(\/'''2''')2, D^2=(1-1/2)^2 +(1/2)=

D^2=1/4-1/2+1=3/4,D=\/'''3''''/2,the shorter,because (\/'''3''' /2)<1

so,(-\/''2'''/2 ,1/2) & (\/''2'''/2 ,1/2) are the points.

God bless you.

Still have questions? Get your answers by asking now.