anonymous
  • anonymous
Using the completing-the-square method, rewrite f(x) = x2 + 4x − 1 in vertex form
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Group the terms like this \[f(x)=(x^2+4x)-1\] Take the coefficient of x, divide it by 2 and square it. What's \(\left( \frac{ 4 }{ 2 } \right)^2\) ?
anonymous
  • anonymous
4
anonymous
  • anonymous
Right, so add 4 inside the parentheses. This means you also have to subtract 4 outside, so the equation doesn't change. \[f(x)=(x^2+4x+4)-1-4\]

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anonymous
  • anonymous
Now factor the part in parentheses, and combine like terms on the outside
anonymous
  • anonymous
so would it be f(x)=(x^2+4x+4)-5
anonymous
  • anonymous
yes, but you also have to factor x² + 4x + 4
anonymous
  • anonymous
oh I got it. But can you help me with a few others?
anonymous
  • anonymous
ok
anonymous
  • anonymous
Using the completing-the-square method, find the vertex of the function f(x) = –2x2 + 12x + 5 and indicate whether it is a minimum or a maximum and at what point.
anonymous
  • anonymous
\(f(x)=-2x^2+12x+5\) Group it again \(f(x)=(-2x^2+12x)+5\) This time, because you have -2 in front of x², you have to factor it out. Can you do that?
anonymous
  • anonymous
would it be f(x)=-2(x^2+12x)+5
anonymous
  • anonymous
no, you have to divide 12 by -2 as well because it's also in parentheses
anonymous
  • anonymous
so it would be 6?
anonymous
  • anonymous
12/(-2)= -6 So you have \[f(x)=-2(x^2-6x)+5\] Now take the coefficient of x, that's -6. Divide it by 2, the square the result like we did above
anonymous
  • anonymous
okay so would it be minimum or maximum?
anonymous
  • anonymous
the number in front of x² is negative, so it's a maximum
anonymous
  • anonymous
okay so whats the point? (-3,5)?
anonymous
  • anonymous
You have to complete the square to find the point
anonymous
  • anonymous
If you don't want to do it that way, use \[x=-\frac{ b }{ 2a }\]
anonymous
  • anonymous
okay I just reset my exam so now I have different questions
anonymous
  • anonymous
ughhhh I need help

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