document.write( "Question 1079111: For the following functions, determine the stationary points and classify them. include graphs.\r
\n" ); document.write( "\n" ); document.write( "a) f(x) = 2x^2 - x + 6
\n" ); document.write( "b) f(x) = X^2 - 4x + 3
\n" ); document.write( "c) f(x) = x^3 - 3x + 3
\n" ); document.write( "d) f(x) = 1 - 9x - 6x^2 -X^3
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Algebra.Com's Answer #693489 by Boreal(15235)\"\" \"About 
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The first two are the vertex of a parabola, and they are minima.
\n" ); document.write( "The first has a vertex at x=-1/4 and y=51/8
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\n" ); document.write( "The vertex is at x=2, y=-1
\n" ); document.write( "\"graph%28300%2C300%2C-10%2C10%2C-10%2C10%2Cx%5E2-4x%2B3%29\"
\n" ); document.write( "This has a derivative of 3x^2-3, so critical points are when 3x^2-3=0, which occurs when x=1 or -1
\n" ); document.write( "When x=-1, y=5 and the second derivative is 6x and negative when x=-1, so that is a local maximum. When x=1, a local minimum, y=1.
\n" ); document.write( "\"graph%28300%2C300%2C-10%2C10%2C-10%2C10%2Cx%5E3-3x%2B3%29\"
\n" ); document.write( "This has a derivative of -3x^2-12x-9, and setting that equal to 0 and factoring out a -3, x^2+4x+3=0, with stationary points at x=-1,-3 and y=5, 1 or (-1, 5) and (-3, 1). Second derivative is -6x-12, so when x=-1 second derivative is negative and a local maximum, and when x=-3, it is positive and a local minimum.
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