SAT Question 841ef26c: Nonlinear Functions answer and explanation with Desmos solution

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