Question 136·200 Super-Hard SAT Math Questions·Advanced Math

A factor is a piece multiplied into $f(x)$, so $f(x)=0$ when that factor is $0$. To find one zero, or input with output $0$, set the given factor to zero:

Type $2x_{1}+3\sim0$. Desmos uses $x_{1}$ as the unknown and $\sim$ to find the value that makes the sides match. Under PARAMETERS, it shows $x_{1}=-1.5=-\frac32$. So the known zero is negative.

The minimum of $g$ occurs at $x=\frac{19}{32}$. In $g(x)=f(4x-1)$, the input of $f$ is the whole expression $4x-1$. Since that expression can reach every input of $f$, the vertex, or lowest point, of $f$ has input

Type $4\left(\frac{19}{32}\right)-1$ in Desmos. It shows $1.375$; the fraction button gives $\frac{11}{8}$. So the axis of symmetry, the vertical mirror line through the vertex, is $f$’s axis $x=\frac{11}{8}$. The given $\frac{19}{32}$ belongs to $g$, not $f$.

A parabola’s zeros are equally far from its axis. To reflect a zero $r$ across $x=h$, double $h$ and subtract $r$. Here $r=-\frac32$, so

Reflect across the vertex’s axis, not across $0$.

Type $2\cdot\frac{11}{8}+\frac32$ in Desmos. It shows $4.25$; the fraction button gives $\frac{17}{4}$. The positive zero of $f$ is $c=\frac{17}{4}$. Choice D.