Question 34·Hard·Advanced Math

The matching outputs $f(0)=f(4)=-19$ put the axis of symmetry, the parabola's mirror line, halfway between $0$ and $4$, at $x=\frac{0+4}{2}=2$. Matching outputs sit equally far from the mirror line. In $a(x-b)^2+c$, that line is $x=b$, so $b=2$.

At $x=2$, the square in $f(x)=a(x-2)^2+c$ is zero, so $f(2)=c=5$. The vertex is the parabola's turning point; its output is $5$, but the question asks for the output at $x=1$.

Use $f(0)=-19$ to find $a$. Put $0$ into $f(x)=a(x-2)^2+5$ and set the output equal to $-19$:

Type $-19\sim a(0-2)^2+5$. The regression symbol $\sim$ asks Desmos to fit the unknown $a$ instead of graphing an equation. Under PARAMETERS, Desmos reports $a=-6$.

Define $f(x)=a(x-2)^2+5$ in Desmos, then type $f(1)$. That means put $1$ in for $x$. Desmos uses its fitted $a=-6$ and prints $-1$, so $f(1)=-1$. Choice C.