SAT Question 95eeeb5b: Nonlinear Functions answer and explanation with Desmos solution

Choice A is correct. It's given that the graph of $y=f\left(x\right)$ in the xy-plane passes through the points $\left(7,0\right)$ and $\left(-3,0\right)$. It follows that when the value of $x$ is either $7$ or $-3$, the value of $f\left(x\right)$ is $0$. It's also given that the function $f$ is defined by $f\left(x\right)=a{x}^2+bx+c$, where $a$, $b$, and $c$ are constants. It follows that the function $f$ is a quadratic function and, therefore, may be written in factored form as $f\left(x\right)=a\left(x-u\right)\left(x-v\right)$, where the value of $f\left(x\right)$ is $0$ when $x$ is either $u$ or $v$. Since the value of $f\left(x\right)$ is $0$ when the value of $x$ is either $7$ or $-3$, and the value of $f\left(x\right)$ is $0$ when the value of $x$ is either $u$ or $v$, it follows that $u$ and $v$ are equal to $7$ and $-3$. Substituting $7$ for $u$ and $-3$ for $v$ in the equation $f\left(x\right)=a\left(x-u\right)\left(x-v\right)$ yields $f\left(x\right)=a\left(x-7\right)\left(x-\left(-3\right)\right)$, or $f\left(x\right)=a\left(x-7\right)\left(x+3\right)$. Distributing the right-hand side of this equation yields $f\left(x\right)=a\left(x^2-7x+3x-21\right)$, or $f\left(x\right)=a{x}^2-4ax-21a$. Since it's given that $f\left(x\right)=a{x}^2+bx+c$, it follows that $b=-4a$. Adding $a$ to each side of this equation yields $a+b=-3a$. Since $a+b=-3a$, if $a$ is an integer, the value of $a+b$ must be a multiple of $3$. If $a$ is an integer greater than $1$, it follows that $a\ge 2$. Therefore, $-3a\le -3\left(2\right)$. It follows that the value of $a+b$ is less than or equal to $-3\left(2\right)$, or $-6$. Of the given choices, only $-6$ is a multiple of $3$ that's less than or equal to $-6$.