SAT Question 415ab1d2: Linear Functions answer and explanation with Desmos solution

Choice A is correct. It's given that the graph of the linear function $y=f\left(x\right)+19$ is shown. This means that the graph of $y=f\left(x\right)+19$ can be translated down $19$ units to create the graph of $y=f\left(x\right)$ and the y-coordinate of every point on the graph of $y=f\left(x\right)+19$ can be decreased by $19$ to find the resulting point on the graph of $y=f\left(x\right)$. The y-intercept of the graph of $y=f\left(x\right)+19$ is $\left(0,3\right)$. Translating the graph of $y=f\left(x\right)+19$ down $19$ units results in a y-intercept of the graph of $y=f\left(x\right)$ at the point $\left(0,3-19\right)$, or $\left(0,-16\right)$. The graph of $y=f\left(x\right)+19$ slants down from left to right, so the slope of the graph is negative. The translation of a linear graph changes its position, but does not change its slope. It follows that the slope of the graph of $y=f\left(x\right)$ is also negative. The equation of a linear function $f$ can be written in the form $f\left(x\right)=b+mx$, where $b$ is the y-coordinate of the y-intercept and $m$ is the slope of the graph of $y=f\left(x\right)$. It's given that $c$ and $d$ are positive constants. Since the y-coordinate of the y-intercept and the slope of the graph of $y=f\left(x\right)$ are both negative, it follows that $f\left(x\right)=-d-cx$ could define $f$.