How do you solve #y=-2x^2+4x+7# using the completing square method?

Question

Hazel Anglin · Accepted Answer

See below.

We take a quadratic equation of the form in order to finish the square. #ax^2+bx+c=0# and transform it into #a(x+d)^2+e=0# Begin by factoring out #-2# to get a coefficient of #1# for the #x^2# term. #y=-2(x^2-2x-7/2)# Now look at the coefficient of the #x# term. #y=-2(x^2color(blue)(-2)x-7/2)# Divide this coefficient by #2# and square the result: #(-2/2)^2=(1)^2=1# I'm going to rewrite the formula: #y=-2(x^2-2x+f-7/2-f)# Replace #f# with the result of the above operation: #=>y=-2(x^2-2x+1-7/2-1)# The first and second parts of the parentheses are divided apart: #=>y=-2[(x^2-2x+1)-7/2-1]# Simplify: #=>y=-2[(x^2-2x+1)-9/2]# A perfect square remains enclosed in parenthesis. Factor: #=>y=-2[(x-1)^2-9/2]# Distribute #-2#: #=>y=-2(x-1)^2+9# Alternatively, equivalently: #y=9-2(x-1)^2#

Savannah Anderson · Answer

undefined To solve the equation $ y = -2x^2 + 4x + 7 $ using the completing the square method:

1. Rewrite the equation in the form $ y = ax^2 + bx + c $.
2. Group the $ x $-terms together and factor out the coefficient of $ x^2 $ from the $ x $-terms.
3. Complete the square by adding and subtracting $(b/2)^2$ inside the parentheses.
4. Rewrite the expression as a perfect square trinomial.
5. Write the expression in vertex form $ y = a(x - h)^2 + k $.
6. Identify the vertex $ (h, k) $ of the parabola.