Question 987386
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{{{f(x)=6x^2+x-2}}}
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Set x=0 to find y-intercept:
{{{f(0)=6(0)^2+0-2}}}
{{{y=-2}}} ANSWER: The y intercept is (0,-2)
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Set y=0 to find x-intercept:
0=6x^2+x-2
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*[invoke quadratic "x", 6, 1, -2 ]
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ANSWER: The x intercepts are (0.5,0) and (-2/3,0)
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Derivatives:
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{{{f}}}'{{{(x)=12x+1}}} Set this to zero to find critical point. 
{{{f}}}''{{{(x)=12}}} This is positive so we will find the minimum. 
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Find  critical point:
Set first derivative=0 to find x value of critical point
{{{f}}}'{{{(x)=12x+1}}}
{{{0=12x+1}}}
{{{-1=12x}}}
{{{-1/12=x}}} 
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Solve for f(-1/12) to find the y value of critical point:
{{{f(x)=6x^2+x-2}}}
{{{y=6(-1/12)^2-1/12-2}}}
{{{y=6/144-12/144-2}}}
y=-2 6/144=-2 1/24
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Critical point (turning point, minumum): (-1/12,-2 1/24)
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GRAPH:
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{{{ graph( 500, 500, -10, 10, -10, 10, 6x^2+x-2) }}}