Question

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

Answer 1

Answer:

y = 2 (x + seven-halves) squared minus one-fourth [y =  $(x + \frac{7}{2} )^{2} - \frac{1}{4}$ ]

Step-by-step explanation:

We know that,

vertex form is y = a(x-h)² + k

vertex is (h, k)

Now,

Given that the equation is -

y = (x+3)² + (x+4)²

 = x² + 3² + 2×3×x + x² + 4² + 2×4×x

= x² + 9 + 6x + x² + 16 + 8x

= 2x² + 14x + 25

= $x^{2} + 7x + \frac{25}{2}$

= $x^{2} + 7x + \frac{25}{2} + \frac{49}{4} - \frac{49}{4}$

= $(x + \frac{7}{2} )^{2} + \frac{25}{2} - \frac{49}{4}$

= $(x + \frac{7}{2} )^{2} - \frac{1}{4}$

∴ we get

The vertex form is -

y = $(x + \frac{7}{2} )^{2} - \frac{1}{4}$

So,

The correct option is - y = 2 (x + seven-halves) squared minus one-fourth (y =  $(x + \frac{7}{2} )^{2} - \frac{1}{4}$ )