Question 357·Hard·Algebra

No Desmos needed. The labeled points give $g$'s outputs directly. A point $(x,y)$ on $g$ means input $x$ produces output $y$. So the two labeled points tell you:

Put $0$ and $4$ into the given rule for $g$:

For example, at $x=0$, the inputs $x+3$ and $x-1$ become $3$ and $-1$. Put the new input into every copy of $f$ before combining its outputs.

Because $f$ is linear, write $f(x)=ax+b$, where $a$ is its slope and $b$ is its output at $0$. Type $f(x)=ax+b$ in Desmos, then type $[2f(3)-f(-1),2f(7)-f(3)]\sim[3,27]$. The $\sim$ asks Desmos to fit both output conditions. Under PARAMETERS, it reports $a=6$ and $b=-39$. The $3$ was $g(0)$, not $f(0)$, so the dial calibration is $f(x)=6x-39$. Choice A.

Let $f(x)=ax+b$, where $a$ is the slope and $b=f(0)$. Substitute that rule into both copies of $f$:

Distribute, including the minus sign before the second bracket:

Combine like terms:

So $g$ keeps the same slope $a$, but its intercept is $7a+b$, not $b$.

The slope is the change in output divided by the change in input. Type $\frac{27-3}{4-0}$ in Desmos; it returns $6$. Since $g(x)=ax+7a+b$, its coefficient of $x$ is $a$, so $a=6$.

The point $(0,3)$ means $g(0)=3$. Using $g(x)=ax+7a+b$ and $a=6$, write:

Type $7(6)+b\sim3$ in Desmos; under PARAMETERS it reports $b=-39$. So the original calibration is $f(x)=6x-39$, not a line with intercept $3$. Choice A.