Question 111497
I'm going to assume that you meant {{{f(x)=-2x^2-6x+5}}}.
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For any quadratic function {{{f(x)=ax^2+bx+c}}} the maximum or minimum occurs at the vertex of the parabola described by the function.  The the x-coordinate of the vertex (h, k) of the parabola {{{ax^2+bx+c}}} is given by {{{h=(-b)/2a}}}.  In this case, {{{-(6/4)}}}.  This is the x value at which the maximum or minimum occurs.
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To find the maximum or minimum value, you need to evaluate {{{f(h)}}}.
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{{{f(-6/4)=-2(-6/4)^2-6(-6/4)+5}}}
{{{f(-6/4)=-2(36/16)+(36/4)+5}}}
{{{f(-6/4)=-72/16+144/16+80/16}}}
{{{f(-6/4)=152/16=38/4}}}
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You didn't say that you needed to identify the value as either a minimum or maximum.  There are a couple of ways to do that.  The easiest is to pick another value for x and evaluate the function at the new value.  If this new value for the function is smaller than the vertex y-coordinate, you know you have found a maximum, and vice versa.  Let's evaluate {{{f(0)=5}}}.  Since 5 (or 20/4) is smaller than 38/4, the point must be a maximum.
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You could also graph the function, and determine whether the vertex was a minimum or maximum by inspection.
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{{{graph(400,400,-10,10,-10,10,-2x^2-6x+5)}}}
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There is yet another way, but you need Calculus.  Write back if you know how to take the first and second derivatives of a function.