Question

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

Answer 1

Answer:

$y=2(x+\frac{7}{2})^2+\frac{1}{2}$ vertex form

Step-by-step explanation:

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

Lets square the terms

$(x+3)^2=x^2+6x+9$

$(x+4)^2= x^2+8x+16$

$y = x^2+6x+9 +x^2+8x+16=2x^2+14x+25$

Now use completing the square method to get the vertex form

to apply the method we need to take out 2

WE take middle term , divide it by 2 and square it . then add and subtract it

$y =2(x^2+7x)+25$

$\frac{7}{2}$

square it $\frac{49}{4}$

Add and subtract the fraction

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

Take out -49/4 and multiply with 2 , then add it with 25

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

$y=2(x^2+7x+\frac{49}{4})+\frac{1}{2}$

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

Answer 2

Answer: B on edge 2020.

Step-by-step explanation: