Question 193·200 Super-Hard SAT Math Questions·Advanced Math

A vertex is a quadratic graph’s turning point. The given vertex $(-2,5)$ makes the squared part zero at $x=-2$, so $f(x)=a(x+2)^2+5$. The multiplier $a$ sets how the parabola opens. Use $x+2$, not $x-2$: the square must be zero at $-2$.

The point $(4,-7)$ means $g(4)=-7$. Put $4$ into the rule for $g$: $-7=g(4)=-6f(4+1)+2=-6f(5)+2$. The point belongs to $g$, so carry its input through $g$ before using $f$. Putting $(4,-7)$ directly into $f$ would use the wrong graph.

Replace $f(5)$ with $a(5+2)^2+5$. Type $-7\sim-6\big(a(5+2)^2+5\big)+2$; the regression symbol $\sim$ tells Desmos to find $a$ from the equation. Under PARAMETERS, Desmos shows $a\approx-0.07142857$. It stores the unrounded value for the next lines.

Type $f(x)=a(x+2)^2+5$, then $f(2)$. Desmos uses its stored $a$ and shows $f(2)\approx3.857142857$; its fraction button gives $\frac{27}{7}$. This is the original function’s output at input $2$.

Type $g(x)=-6f(x+1)+2$, then $g(2)$. Desmos shows $g(2)\approx-17.285714286$; its fraction button gives $-\frac{121}{7}$. Here the input to $f$ is $3$, because $2+1=3$.

Type $g(2)-f(2)$. Desmos shows about $-21.142857143$, and its fraction button gives $-\frac{148}{7}$. Subtract $f(2)$ rather than adding it: the requested value of $g(2)-f(2)$ is $-\frac{148}{7}$. Choice A.