Calculus
posted by Elizabeth .
Find the point on the graph of function that is closest to the point.
f(x)=x^2 (2,1/2)

From the given point P0(x0,y0)=(2,1/2), we draw a line to join a point on the curve
f(x)=x^2 at P(x,y).
The distance PP0 will be shortest when PP0 is perpendicular to the tangent of f(x) at P.
The slope of PP0 is given by:
m1=(y0y)/(x0x)
=(y0x^2)/(x0x) since P is on curve f(x)=x^2.
The tangent of the curve f(x) at P is given by
m2=f'(x)=2x
Perpendicularity requires that m1*m2=1, or
[(y0x^2)/(x0x)]*(2x)=1
which simplifies to a cubic equation
x³1=0 which factorizes easily to
(x1)(x²+x+1)=0
The first factor gives x=1, and the second factor results in two complex roots which do not concern us.
Therefore the required point is P(1,1), or P(1,1) is the point on f(x) that is closest to the given point (1, 1/2). 
Thanks!

You're welcome!