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\(m = 4 - (4 - x)^2\)

For m to be greatest : \((4 - x)^2 = 0\)

Answer D
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Asked: If \(m = 4-(4-x)^2\), what is the greatest possible value of m?

Greatest value of m = 4 when x=4

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Explode the equation to bring it to a second degree equation

We see that y= 4-(4-x)^2 = -x^2+8x-12

Now we have the equation of a parable (y=ax^2+bx+c) and we know that a parable takes a U form for a>0 (thus its vertex is the minimum point of y for Xv which is the x coordinate of the vertex) and that a parable directs downward if a<0 (thus the vertex becomes the maximum point of the function and Xv is the variable for which Y takes the max value).

Thus we calculate the X that maximize the y value with the formula (-b/2a) which is the x coordinate of the vertex of the parable so xv=-8/-2=4

Then substitute x with 4 and get the max value
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