How do you write #F(x) = -x^2+4x# in vertex form?

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Answer 1 · 2016-05-20T11:15:30

#y=-(x-2)^2+4 -> g(x) =-(x-2)^2+4#

Standard form #y=ax^2+bx+c#
Vertex form #y=a(x+b/(2a))^2+k+c#

Where #a(b/(2a))^2+k=0#

#k# neutralizes the error introduced
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Step 1 #" "-1(x^(color(magenta)(2))-4x)+0#

Step 2#" "-1(x-4x)^(color(magenta)(2))+color(magenta)(k)+0#

Step 3#" "-1(x-4)^2+k" "larr" removed the "x" from "4x#

Step 4#" "-1(x-2)^2+k" "larr" halved the 4"#

Step 4#" "k+[-1(-2)^2] =0 -> k=+4#

Step 5#" "-(x-2)^2+4#

Tony B