document.write( "Question 462469: Using the formula f(x) = x^2 – 2x + 1, find if it has a maximum or minimum and give that point. Also give x-intercepts. \n" ); document.write( "
Algebra.Com's Answer #317040 by nerdybill(7384)\"\" \"About 
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Using the formula f(x) = x^2 – 2x + 1, find if it has a maximum or minimum and give that point. Also give x-intercepts.
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\n" ); document.write( "f(x) = x^2 – 2x + 1
\n" ); document.write( "Examining the coefficient associated with the x^2 term, we see that it is \"positive\" -- this means the vertex is a MINIMUM.
\n" ); document.write( "x = -b/(2a)
\n" ); document.write( "x = -(-2)/(2*1)
\n" ); document.write( "x = (2)/(2)
\n" ); document.write( "x = 1
\n" ); document.write( "To find the 'y', plug above value back into:
\n" ); document.write( "f(x) = x^2 – 2x + 1
\n" ); document.write( "f(1) = 1^2 – 2(1) + 1
\n" ); document.write( "f(1) = 1 – 2 + 1
\n" ); document.write( "f(1) = 0
\n" ); document.write( "Vertex: (1,0)
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\n" ); document.write( "x-intercepts: set f(x) to zero and solve for x:
\n" ); document.write( "f(x) = x^2 – 2x + 1
\n" ); document.write( "0 = x^2 – 2x + 1
\n" ); document.write( "0 = (x-1)(x-1)
\n" ); document.write( "x = 1
\n" ); document.write( "x-intercept at (1,0) (same as the vertex, in this case)
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