anonymous
  • anonymous
A function is defined as f(x)=ax^2+bx+d, where a, b, and d are integers. The minimum value of f(x) is -4 Determine the x-coordinate of the minimum value of f(x) and hence find the value of d.
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
okay, well you know that at some x the minimum is -4, so you can find that point by taking the derivative of f(x) and setting it to 0 which is: \[2ax +b = 0\] so \[x = -b/2a\] then you can place that in the first equation: \[ax^2 + bx + d = a(-b/2a)^2 + b(-b/2a) + d = b^2/2a -b^2/2a + d = -4\] thus \[d = -4\] then you can find x by setting f(x) = -4 and then solving for x, which gives \[-4 = ax^2 +bx -4 --> 0 = ax^2+bx = x(ax + b)\] so you can say that \[x = 0, -b/a\]
anonymous
  • anonymous
sorry haha, math error :O. just noticed. should be \[b^2/4a^2 - b^2/2a + d = -4\] which would change everything. its kinda late here......sorry.
anonymous
  • anonymous
so then \[x = -b/2a\] and \[d = (-16a^2 +b^2(2a-1))/4a^2\]

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anonymous
  • anonymous
not \[4a^2\] i meant \[4/a\]....I am going to bed.
anonymous
  • anonymous
So to actually put it together nicely. \[x=-b/2a\] and \[d = -4 - b^2/4a + b^2/2a = -4 + b^2/4a\].

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