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How do you complete the square for #y=x^2 - 9x#?

↳Redirected from "How do I change #int_0^1int_0^sqrt(1-x^2)int_sqrt(x^2+y^2)^sqrt(2-x^2-y^2)xydzdydx# to cylindrical or spherical coordinates?"

1 Answer

Jul 10, 2015

#y=(x-9/2)^2-81/4#

Explanation:

#(x+a)^2=x^2+2ax+a^2# => #x^2+bx=(x+b/2)^2-b^2/4#
Therefore substituting #b=-9# we have:
#y=x^2-9x=(x-9/2)^2-81/4#

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