How do you rewrite #y=(x+3)^2+(x+4)^2# in vertex form?

1 Answer
Jim G.
May 9, 2017

#y=2(x+7/2)^2+1/2#

Explanation:

The equation of a parabola in #color(blue)"vertex form"# is.

#color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))#
where ( h , k ) are the coordinates of the vertex and a is a constant.

#"Expand and simplify y"#

#y=x^2+6x+9+x^2+8x+16#

#color(white)(y)=2x^2+14x+25#

#"using the method of "color(blue)"completing the square"#

#y=2(x^2+7x)+25larr" coefficient of " x^2" term is unity"#

add #(1/2"coefficient of x-term")^2" to " x^2+7x#

#"we must also subtract this value"#

#"that is add/subtract " (7/2)^2=49/4#

#y=2(x^2+7xcolor(red)(+49/4)color(red)(-49/4))+25#

#y=2(x+7/2)^2-49/2+25#

#rArry=2(x+7/2)^2+1/2larrcolor(red)" in vertex form"#

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