What is the first step when rewriting #y=-4x^2+2x-7# in the form #y=a(x-h)^2+k#?

3 Answers
Jul 3, 2017

There is a process for completing the square but the values, #a,h, and k# are far too easy to obtain by other methods. Please see the explanation.

Explanation:

  1. #a = -4# the value of "a" is always the leading coefficient of the #x^2# term.
  2. #h=-b/(2a) = -2/(2(-4)) = 1/4#
  3. #k = y(h) = y(1/4) = -4(1/4)^2+2(1/4)-7 = -27/4#

This is a lot easier than adding zero to the original equation in the form of #-4h^2+4h^2#:

#y = -4x^2+2x-4h^2+4h^2-7#

Removing a factor of -4 from the first 3 terms:

#y = -4(x^2-1/2x+h^2)+4h^2-7#

Match the middle term of the expansion #(x-h)^2=x^2-2hx+h^2# with the middle term in the parenthesis:

#-2hx = -1/2x#

Solve for h:

#h = 1/4#

Therefore, we can compress the 3 terms into #(x-1/4)^2#:

#y = -4(x-1/4)^2+4h^2-7#

Substitute for h:

#y = -4(x-1/4)^2+4(1/4)^2-7#

Combine like terms:

#y = -4(x-1/4)^2-27/4#

Look at how much easier is it to remember 3 simple facts.

Jul 3, 2017

You would factor out the #-4# from the first term giving you
#y=-4(x^2-1/2x)-7#

Explanation:

First complete the square.
#y=-4x^2+2x-7#
get the #x^2# term to have a coefficient of #1#.
You can do this by factoring out #-4# from the first two terms.
#y=-4(x^2-1/2x)-7#
Then complete the square
#y=-4(x-1/4)^2-7-(1/16xx-4)#

this simplifies down to
#y=-4(x-1/4)^2-6.75#

Jul 3, 2017

Factor out #-4# from each term, to get:

#y = -4[x^2-1/2x+7/4]#

Explanation:

#y = ax^2 +bx+c#

In order to complete the square, the coefficient of #x^2# must be #1#, so the first step will be to make this happen.

#y = -4x^2 +2x-7" "larr# factor out #-4# from each term to get:

#y = -4[x^2-1/2x+7/4]#

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For the sake of completeness the full process is shown below.
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#color(blue)(y = -4[x^2-1/2x" "+7/4])" "larr# add and subtract #(b/2)^2#

#b= -1/2" "rArr color(red)((b/2)^2 = (-1/2 div 2)^2 =(-1/4)^2 = 1/16)#

#color(blue)(y = -4[x^2-1/2x color(red)(+1/16 - 1/16)color(blue)(+7/4)])#

#y = -4[(x^2-1/2x +1/16)+( - 1/16+7/4)]#

#y = -4[(x-1/4)^2 +27/16]" "larr# distribute the #-4#

#y = -4(x-1/4)^2 -27/4#

#y = -4(x-1/4)^2 - 6 3/4#