SAT Question 92f812bb: Nonlinear Functions answer and explanation with Desmos solution

Choice D is correct. The equation of a parabola in the xy-plane can be written in the form $y=a{\left(x-h\right)}^2+k$, where $a$ is a constant and $\left(h,k\right)$ is the vertex of the parabola. If $a$ is positive, the parabola will open upward, and if $a$ is negative, the parabola will open downward. It's given that the parabola has vertex $\left(9,-14\right)$. Substituting $9$ for $h$ and $-14$ for $k$ in the equation $y=a{\left(x-h\right)}^2+k$ gives $y=a{\left(x-9\right)}^2-14$, which can be rewritten as $y=a\left(x-9\right)\left(x-9\right)-14$, or $y=a\left(x^2-18x+81\right)-14$. Distributing the factor of $a$ on the right-hand side of this equation yields $y=a{x}^2-18ax+81a-14$. Therefore, the equation of the parabola, $y=a{x}^2-18ax+81a-14$, can be written in the form $y=a{x}^2+bx+c$, where $a=a$, $b=-18a$, and $c=81a-14$. Substituting $-18a$ for $b$ and $81a-14$ for $c$ in the expression $a+b+c$ yields $\left(a\right)+\left(-18a\right)+\left(81a-14\right)$, or $64a-14$. Since the vertex of the parabola, $\left(9,-14\right)$, is below the x-axis, and it's given that the parabola intersects the x-axis at two points, the parabola must open upward. Therefore, the constant $a$ must have a positive value. Setting the expression $64a-14$ equal to the value in choice D yields $64a-14=-12$. Adding $14$ to both sides of this equation yields $64a=2$. Dividing both sides of this equation by $64$ yields $a=\frac{2}{64}$, which is a positive value. Therefore, if the equation of the parabola is written in the form $y=a{x}^2+bx+c$, where $a$, $b$, and $c$ are constants, the value of $a+b+c$ could be $-12$.