SAT Question 7bf77c19: Nonlinear Functions answer and explanation with Desmos solution

Choice D is correct. It’s given that $a$ and $b$ are integer constants such that $a<b$ and $b<0$. The y-coordinate of the y-intercept of the graph of a function is the value of the function when $x=0$. Equation I is $p\left(x\right)=a(4.1{)}^x+b$. Substituting $0$ for $x$ in this equation yields $p\left(0\right)=a(4.1{)}^0+b$, which can be rewritten as $p(0)=a(1)+b$, or $p(0)=a+b$. Therefore, the y-coordinate of the y-intercept of the graph of $y=p(x)$ is $a+b$. Equation I has a coefficient of $a$ and a constant $b$. Thus, the value of the y-coordinate of the y-intercept, $a+b$, is not displayed as a constant or coefficient in equation I. Equation II is $r\left(x\right)=a(4.1{)}^{x+b}$. Similarly, substituting $0$ for $x$ in this equation yields $r\left(0\right)=a(4.1{)}^{0+b}$, which can be rewritten as $r\left(0\right)=a(4.1{)}^b$. Since $b<0$, the value of $(4.1{)}^b$ is not equal to $1$. Thus, the value of $a(4.1{)}^b$ is not equal to $a$. The value of $a(4.1{)}^b$ is also not equal to $b$, since there is no pair of integers $a$ and $b$ such that $a(4.1{)}^b=b$. Thus, the value of the y-coordinate of the y-intercept for this equation is $a(4.1{)}^b$, which is not displayed in equation II as a constant or coefficient. The coefficient in equation II is $a$, and the only constant, $b$, appears in the exponent. Therefore, neither equation I nor equation II displays, as a constant or coefficient, the y-coordinate of the y-intercept of the graph of the corresponding function.