So circles really are special cases of ellipses 2007

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Unformatted text preview: to do this step. Now, this is where the process really starts differing from what we’ve seen to this point. We still take one-half the coefficient of x and square it. However, instead of adding this to both sides we do the following with it. © 2007 Paul Dawkins 214 http://tutorial.math.lamar.edu/terms.aspx College Algebra 2 2 æ 6ö ç - ÷ = ( -3 ) = 9 è 2ø 3ö æ f ( x ) = 2 ç x2 - 6 x + 9 - 9 + ÷ 2ø è We add and subtract this quantity inside the parenthesis as show. Note that all we are really doing here is adding in zero since 9-9=0! The order listed here is important. We MUST add first and then subtract. The next step is to factor the first three terms and combine the last two as follows. 15 ö 2 æ f ( x ) = 2 ç ( x - 3) - ÷ 2ø è As a final step we multiply the 2 back through. f ( x ) = 2 ( x - 3) - 15 2 And there we go. [Return to Problems] (b) Be careful here. We don’t have a coefficient of 1 on the x2 term, we’ve got a coefficient of -1. So, the process is identical outside of that so we won’t put in as muc...
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