document.write( "Question 528114: which point of the graph f(x)=x^2 are the closest to the point (2, 0.5)? \n" ); document.write( "

Algebra.Com's Answer #349409 by htmentor(1343) $\"\"$ $\"About$

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which point of the graph f(x)=x^2 are the closest to the point (2, 0.5)?
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\n" ); document.write( "One way to solve is to find the slope of the tangent line.
\n" ); document.write( "The line passing through the point (2, 0.5) and having the minimum distance to
\n" ); document.write( "f(x) will perpendicular to the tangent line (normal line).
\n" ); document.write( "Let the point on the graph = (x0,y0)
\n" ); document.write( "The slope of the tangent line = df(x)/dx = 2x = 2x0 at the point x0
\n" ); document.write( "The normal line is perpendicular to the tangent line, so its slope is the neg. reciprocal:
\n" ); document.write( "Slope(normal) = -1/(2x0)
\n" ); document.write( "But the slope of a line going through the points (x0,y0) and (2,0.5) is:
\n" ); document.write( "(y0-0.5)/(x0-2)
\n" ); document.write( "We can equate the two expressions for the slope:
\n" ); document.write( "-1/(2x0) = (y0-0.5)/(x0-2)
\n" ); document.write( "But y0 = x0^2 from the formula for the graph:
\n" ); document.write( "Substitute this value and cross-multiply to solve for x0:
\n" ); document.write( "x0^3 = 1 -> x0 = 1
\n" ); document.write( "And since y0 = x0^2 -> y0 = 1
\n" ); document.write( "So the point is (1,1) \n" ); document.write( "

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