SAT Question 97e50fa2: Nonlinear Equations In One Variable And Systems Of Equations In Two Variables answer and explanation with Desmos solution

Substituting $x=a$ in the definitions for $f$ and $g$ gives $f(a)=-\frac{1}{2}(a-4{)}^2+10$ and $g(a)=-a+10$, respectively. If $f(a)=g(a)$, then $-\frac{1}{2}(a-4{)}^2+10=-a+10$. Subtracting $10$ from both sides of this equation gives $-\frac{1}{2}(a-4{)}^2=-a$. Multiplying both sides by $-2$ gives $(a-4{)}^2=2a$. Expanding $(a-4{)}^2$ gives $a^2-8a+16=2a$. Combining the like terms on one side of the equation gives $a^2-10a+16=0$. One way to solve this equation is to factor $a^2-10a+16$ by identifying two numbers with a sum of $-10$ and a product of $16$. These numbers are $-2$ and $-8$, so the quadratic equation can be factored as $(a-2)(a-8)=0$. Therefore, the possible values of $a$ are either $2$ or $8$.

Alternate approach: Graphically, the condition $f(a)=g(a)$ implies the graphs of the functions $y=f(x)$ and $y=g(x)$ intersect at $x=a$. The graph $y=f(x)$ is given, and the graph of $y=g(x)$ may be sketched as a line with $y$-intercept $10$ and a slope of $-1$ (taking care to note the different scales on each axis). These two graphs intersect at $x=2$ and $x=8$. Note that 2 and 8 are examples of ways to enter a correct answer.