How do you solve #y=-(x-4)^2+1#?

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Answer 1 · 2015-07-10T20:49:52

The solution is #y=x^2-8+17#.

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

#(x-4)^2# represents a square of a difference, the formula of which is #(x-4)^2=a^2+2ab+b^2#, where #a=x,# and #b=-4#.

#(x-4)^2=(x^2)+(2*x*-4)+(-4)^2# =

#x^2-8x+16#

Substitute the equation back to the original equation.

#y=x^2-8x+16+1#=

#y=x^2-8+17#