How do you rewrite #y=(x+3)^2+(x+4)^2# in vertex form?
Risposta:
#y=2(x+7/2)^2+1/2#
Spiegazione:
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"#