document.write( "Question 308022: what is the maximum or minimum value of the quardratic function ?\r
\n" ); document.write( "\n" ); document.write( "f(x)=3x^2+12x+8\r
\n" ); document.write( "\n" ); document.write( "a. -28\r
\n" ); document.write( "\n" ); document.write( "b. -4\r
\n" ); document.write( "\n" ); document.write( "c. -2\r
\n" ); document.write( "\n" ); document.write( "d. 44\r
\n" ); document.write( "\n" ); document.write( "tthanks!
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Algebra.Com's Answer #220263 by Fombitz(32388)\"\" \"About 
You can put this solution on YOUR website!
You can either differentiate and find when the derivative is zero.
\n" ); document.write( "f(x)=\"3x%5E2%2B12x%2B8+\"
\n" ); document.write( "f'(x)=\"6x%2B12=0+\"
\n" ); document.write( "\"x=-2+\"
\n" ); document.write( "\"f%28-2%29=3%284%29%2B12%28-2%29%2B8+\"
\n" ); document.write( "\"f%28-2%29=12-24%2B8=-4+\"
\n" ); document.write( "Use the second derivative to find whether its min or max.
\n" ); document.write( "f''(x)=\"6%3E0+\"
\n" ); document.write( "Since the second derivative is positive, the value is a minimum.
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\n" ); document.write( "Or algebraically complete the square to find the vertex form,
\n" ); document.write( "\"f%28x%29=3x%5E2%2B12x%2B8+\"
\n" ); document.write( "\"f%28x%29=3%28x%5E2%2B4x%29%2B8+\"
\n" ); document.write( "\"f%28x%29=3%28x%5E2%2B4x%2B4%29%2B8-12+\"
\n" ); document.write( " \"+f%28x%29=3%28x%2B2%29%5E2-4+\"\r
\n" ); document.write( "\n" ); document.write( "Since the \"x%5E2\" coefficient is positive (3), the parabola opens upwards and the value at the vertex is a minimum and occurs at (-2,-4).
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\n" ); document.write( "b) is the correct answer.
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