SAT Question 781c2f6e: Nonlinear Functions answer and explanation with Desmos solution

Choice D is correct. A y-intercept of a graph in the xy-plane is a point where the graph intersects the y-axis, or a point where $x=0$. Substituting $0$ for $x$ in the equation defining function $f$ yields $f\left(0\right)=a\left(2.2^0+2.2^b\right)$, or $f\left(0\right)=a\left(1+2.2^b\right)$. So, the y-coordinate of the y-intercept of the graph is $a\left(1+2.2^b\right)$, or equivalently, $a+a{\left(2.2\right)}^b$. It's given that function $g$ is equivalent to function $f$, where $0<a<b$. It follows that $k=2.2^b$. Since $a{\left(2.2\right)}^b$ can't be equal to $0$, the coefficient $a$ can't be equal to $a+a{\left(2.2\right)}^b$. Since $0<a$, the constant $k$, which is equal to $2.2^b$, can't be equal to $a+a{\left(2.2\right)}^b$. Therefore, function $g$ doesn't display the y-coordinate of the y-intercept of the graph of $y=f\left(x\right)$ in the xy-plane as a constant or coefficient. It's also given that function $h$ is equivalent to function $f$, where $0<a<b$. The equation defining $f$ can be rewritten as $f\left(x\right)=a{\left(2.2\right)}^x+a{\left(2.2\right)}^b$. It follows that $m=a{\left(2.2\right)}^b$. Since $a{\left(2.2\right)}^b$ can't be equal to $0$, the coefficient $a$ can't be equal to $a+a{\left(2.2\right)}^b$. Since $0<a$, the constant $m$, which is equal to $a{\left(2.2\right)}^b$, can't be equal to $a+a{\left(2.2\right)}^b$. Therefore, function $h$ doesn't display the y-coordinate of the y-intercept of the graph of $y=f\left(x\right)$ in the xy-plane as a constant or coefficient. Thus, neither function $g$ nor function $h$ displays the y-coordinate of the y-intercept of the graph of $y=f\left(x\right)$ in the xy-plane as a constant or coefficient.