Answered

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

y = 2(x + 7)2 – 73

y = (x + 7)2 – 24

y=2(x+(7/2))^2-(1/4)

y=2(x+(7/2))^2+(1/2)

Verified Answer :

Nirina7

y = (x + 3)2 + (x + 4)2= x²+9+6x+x²+16+8x=2x²+14x+25
y=2x²+14x+25
the vertex form of y=ax²+bx+c is y=a(x-h)²+k
where h= -b/2a
and k= c-b²/4a
so the vertex form is y=2(x-(-14/4))²+ 25-196/8=2(x+14/4)²+4/8=2(x+7/2)²+1/2
the answer is y=2(x+(7/2))^2+(1/2)

Answer:

The correct option is 4.

Step-by-step explanation:

The given equation is

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

Simplify the above equation.

\(y=x^2+6x+9+x^2+8x+16\)       \([\because (a+b)^2=a^2+2ab+b^2]\)

\(y=2x^2+14x+25\)

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

Add and subtract the value of  \((\frac{-b}{2a})^2\) in the above parenthesis.

\((\frac{-b}{2a})^2=(\frac{-7}{2(1)})^2=(\frac{7}{2})^2\)

\(y=2[x^2+7x+(\frac{7}{2})^2-(\frac{7}{2})^2]+25\)

\(y=2[x+(\frac{7}{2})]^2-2[(\frac{7}{2})^2]+25\)        \([\because (a+b)^2=a^2+2ab+b^2]\)

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

\(y=2(x+\frac{7}{2})^2+\frac{50-49}{2}\)

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

Therefore correct option is 4.

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