SAT Question 1178f2df: Nonlinear Functions answer and explanation with Desmos solution

It's given that $f$ is a quadratic function. It follows that $f$ can be defined by an equation of the form $f\left(x\right)=a{\left(x-h\right)}^2+k$, where $a$, $h$, and $k$ are constants. It's also given that the table shows three values of $x$ and their corresponding values of $y$, where $y=f\left(x\right)+4$. Substituting $a{\left(x-h\right)}^2+k$ for $f\left(x\right)$ in this equation yields $y=a{\left(x-h\right)}^2+k+4$. This equation represents a quadratic relationship between $x$ and $y$, where $k+4$ is either the maximum or the minimum value of $y$, which occurs when $x=h$. For quadratic relationships between $x$ and $y$, the maximum or minimum value of $y$ occurs at the value of $x$ halfway between any two values of $x$ that have the same corresponding value of $y$. The table shows that x-values of $21$ and $25$ correspond to the same y-value, $-8$. Since $23$ is halfway between $21$ and $25$, the maximum or minimum value of $y$ occurs at an x-value of $23$. The table shows that when $x=23$, $y=8$. It follows that $h=23$ and $k+4=8$. Subtracting $4$ from both sides of the equation $k+4=8$ yields $k=4$. Substituting $23$ for $h$ and $4$ for $k$ in the equation $y=a{\left(x-h\right)}^2+k+4$ yields $y=a{\left(x-23\right)}^2+4+4$, or $y=a{\left(x-23\right)}^2+8$. The value of $a$ can be found by substituting any x-value and its corresponding y-value for $x$ and $y$, respectively, in this equation. Substituting $25$ for $x$ and $-8$ for $y$ in this equation yields $-8=a{\left(25-23\right)}^2+8$, or $-8=a{\left(2\right)}^2+8$. Subtracting $8$ from both sides of this equation yields $-16=a{\left(2\right)}^2$, or $-16=4a$. Dividing both sides of this equation by $4$ yields $-4=a$. Substituting $-4$ for $a$, $23$ for $h$, and $4$ for $k$ in the equation $f\left(x\right)=a{\left(x-h\right)}^2+k$ yields $f\left(x\right)=-4{\left(x-23\right)}^2+4$. The y-intercept of the graph of $y=f\left(x\right)$ in the xy-plane is the point on the graph where $x=0$. Substituting $0$ for $x$ in the equation $f\left(x\right)=-4{\left(x-23\right)}^2+4$ yields $f\left(0\right)=-4{\left(0-23\right)}^2+4$, or $f\left(0\right)=-4{\left(-23\right)}^2+4$. This is equivalent to $f\left(0\right)=-2,112$, so the y-intercept of the graph of $y=f\left(x\right)$ in the xy-plane is $\left(0,-2,112\right)$. Thus, the y-coordinate of the y-intercept of the graph of $y=f\left(x\right)$ in the xy-plane is $-2,112$.