SAT Question ce508fb0: Nonlinear Functions answer and explanation with Desmos solution

9$.Thevalueof$f \left( - 4 \right)$isthevalueof$f \left( x \right)$when$x = - 4$.Substituting$- 4$for$x$intheequation$f \left( x \right) = 3 + \left| - 2 x - x^{2} \right|$yields$f \left( - 4 \right) = 3 + \left| - 2 \left( - 4 \right) - \left( - 4 \right)^{2} \right|$,or$f \left( - 4 \right) = 3 + \left| - 8 \right|$,whichisequivalentto$f \left( - 4 \right) = 3 + 8$,or$f \left( - 4 \right) = 11$.Sincei{t}^{\prime }sgiventhat$f \left( - 4 \right) = c$,itfollowsthat$c = 11$andthevalueof$g \left( c \right)$isthevalueof$g \left( 11 \right)$.Substituting＄11$ for $w$ in the equation $g\left(w\right)=\left|\frac{-w}{w-1}\right|-w+5$ yields $g\left(11\right)=\left|\frac{-11}{11-1}\right|-11+5$, or $g\left(11\right)=\left|-1.1\right|-6$, which is equivalent to $g\left(11\right)=1.1-6$, or $g\left(11\right)=-4.9$. Note that -49/10, -4.9, and -4.90 are examples of ways to enter a correct answer.