Question 721·Medium·Math

The line crosses the $y$-axis at $(0,-3)$ and the $x$-axis at $(1,0)$. Those give outputs at two different inputs, enough to determine a linear function. But the drawn outputs belong to $3f(x)-6$, not directly to $f(x)$.

At each point, put its $x$-coordinate into the graphed equation and set the result equal to its $y$-coordinate:

Undo the changes outside $f$ before treating a graphed output as a value of $f$.

Write the linear function as $f(x)=ax+b$, where $a$ is its slope and $b=f(0)$. Then $f(1)=a+b$. Type $[3b-6,3(a+b)-6]\sim[-3,0]$ in Desmos; the tilde asks it to fit both equations at once. Under PARAMETERS, Desmos reports $a=1$ and $b=1$. So the function asked for is $f(x)=x+1$. Choice A.

Use $(0,-3)$ and $(1,0)$. The slope is the change in output divided by the change in input:

Type $\frac{0-(-3)}{1-0}$ in Desmos to evaluate the fraction; it shows $3$:

This is the slope of the drawn line, not yet the slope of $f$.

The drawn line crosses the $y$-axis at $-3$. In slope-intercept form, $y=mx+b$, that makes $b=-3$. Use the slope you found:

The graph's equation and the given equation describe the same line, so set their outputs equal:

Add $6$ to both sides:

Divide both sides by $3$:

That's the rule for $f$, rather than for its transformed graph. Choice A.