SAT Question 4d037075: Nonlinear Functions answer and explanation with Desmos solution

Choice C is correct. It's given that $f\left(x\right)=\frac{a}{x+b}$ and that the graph shown is a partial graph of $y=f\left(x\right)$. Substituting $y$ for $f\left(x\right)$ in the equation $f\left(x\right)=\frac{a}{x+b}$ yields $y=\frac{a}{x+b}$. The graph passes through the point $\left(-7,-2\right)$. Substituting $-7$ for $x$ and $-2$ for $y$ in the equation $y=\frac{a}{x+b}$ yields $-2=\frac{a}{-7+b}$. Multiplying each side of this equation by $-7+b$ yields $-2\left(-7+b\right)=a$, or $14-2b=a$. The graph also passes through the point $\left(-5,-6\right)$. Substituting $-5$ for $x$ and $-6$ for $y$ in the equation $y=\frac{a}{x+b}$ yields $-6=\frac{a}{-5+b}$. Multiplying each side of this equation by $-5+b$ yields $-6\left(-5+b\right)=a$, or $30-6b=a$. Substituting $14-2b$ for $a$ in this equation yields $30-6b=14-2b$. Adding $6b$ to each side of this equation yields $30=14+4b$. Subtracting $14$ from each side of this equation yields $16=4b$. Dividing each side of this equation by $4$ yields $4=b$. Substituting $4$ for $b$ in the equation $14-2b=a$ yields $14-2\left(4\right)=a$, or $6=a$. Substituting $6$ for $a$ and $4$ for $b$ in the equation $f\left(x\right)=\frac{a}{x+b}$ yields $f\left(x\right)=\frac{6}{x+4}$. It's given that $g\left(x\right)=f\left(x+4\right)$. Substituting $x+4$ for $x$ in the equation $f\left(x\right)=\frac{6}{x+4}$ yields $f\left(x+4\right)=\frac{6}{x+4+4}$, which is equivalent to $f\left(x+4\right)=\frac{6}{x+8}$. It follows that $g\left(x\right)=\frac{6}{x+8}$.