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 Tony B

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