# 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"