Question

Which equation is y = (x + 3)2 + (x + 4)2 rewritten in vertex form?

Answer 1

we have

$y=(x+3)^{2}+(x+4)^{2}$

$y=(x+3)^{2}+(x+4)^{2} \\y= x^{2} +6x+9+x^{2}+8x+16 \\y=2x^{2}+14x+25$

rewrite now in vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$y-25=2x^{2}+14x$

Factor the leading coefficient

$y-25=2(x^{2}+7x)$

Complete the square. Remember to balance the equation by adding the same constants to each side

$y-25+24.50=2(x^{2}+7x+3.5^{2})$

$y-0.50=2(x^{2}+7x+12.25)$

Rewrite as perfect squares

$y-0.50=2(x+3.5)^{2}$

$y=2(x+3.5)^{2}+0.50$

the vertex is the point $(-3.5,0.50)$

therefore

the answer is

$y=2(x+3.5)^{2}+0.50$

Answer 2

y=(x+3)^2+(x+4)^2

First expand to a quadratic

y=2x^2+14x+25

next divide by the leading coefficient (2), or rather factor out the leading coefficient:

y=2[x^2+7x+12.5]]

Complete squares on the expression inside brackets

y=2[(x+3.5)^2-12.25+12.5]

=2[(x+3.5)^2+0.25]

=2[(x+3.5)^2+0.25]

Distribute:

y=2(x+3.5)^2+0.5

The last line above is the vertex form of the same parabola.