Nonlinear functions SAT question

Choice A is correct. It's given that $g(x)=f(x+5)$. Since $f(x)=4x^2+64x+262$, it follows that $f(x+5)=4{(x+5)}^2+64(x+5)+262$. Expanding the quantity ${(x+5)}^2$ in this equation yields $f(x+5)=4(x^2+10x+25)+64(x+5)+262$. Distributing the $4$ and the $64$ yields $f(x+5)=4x^2+40x+100+64x+320+262$. Combining like terms yields $f(x+5)=4x^2+104x+682$. Therefore, $g(x)=4x^2+104x+682$. For a quadratic function defined by an equation of the form $g(x)=a{(x-h)}^2+k$, where $a$, $h$, and $k$ are constants and $a$ is positive, $g(x)$ reaches its minimum, $k$, when the value of $x$ is $h$. The equation $g(x)=4x^2+104x+682$ can be rewritten in this form by completing the square. This equation is equivalent to $g(x)=4(x^2+26)+682$, or $g(x)=4(x^2+26x+169-169)+682$. This equation can be rewritten as $g(x)=4({(x+13)}^2-169)+682$, or $g(x)=4{(x+13)}^2-4(169)+682$, which is equivalent to $g(x)=4{(x+13)}^2+6$. This equation is in the form $g(x)=a{(x-h)}^2+k$, where $a=4$, $h=-13$, and $k=6$. Therefore, $g(x)$ reaches its minimum when the value of $x$ is $-13$.

Choice B is incorrect. This is the value of $x$ for which $f(x)$, rather than $g(x)$, reaches its minimum.

Choice C is incorrect and may result from conceptual or calculation errors.

Choice D is incorrect. This is the value of $x$ for which $f(x-5)$, rather than $f(x+5)$, reaches its minimum.