Once you have a formula for $f(x)$ in terms of $g(x)$, plug in each $x$ from the table to find $f(-27)$, $f(-9)$, and $f(21)$. Those become three points on the graph of $y=f(x)$.

Because $f$ is linear, it has the form $f(x)=mx+b$. Use any two of your points to find the slope $m$, then plug one point into $y=mx+b$ to solve for $b$.

Remember that the $y$-intercept of $y=f(x)$ is the point where $x=0$, which corresponds to the constant term $b$ in $f(x)=mx+b$.

Create a table in Desmos with one column labeled $x_1$ and another labeled $g_1$. Enter the three pairs from the problem: $(-27,3)$, $(-9,0)$, and $(21,5)$.

Add a third column to the same table and enter the formula g1*(x1+3) in its header. This column now shows $f(x_1)$ for each $x_1$ using $f(x)=g(x)(x+3)$.

Create a new table with $x_2$ equal to the $x_1$ values and $y_2$ equal to the $f(x_1)$ values you just computed (you can retype them or copy). Then, in a new expression line, type y2 ~ m x2 + b to perform a linear regression and find the line $y = mx + b$ that matches $f(x)$.

Look at the regression parameters $m$ and $b$ that Desmos reports. The value of $b$ is the $y$-intercept of $f(x)$. Choose the answer option whose $y$-coordinate matches this $b$ value.

We are given

so we can rewrite this as

Use the table values of $x$ and $g(x)$ to find $f(x)$:

• For $x=-27$: $f(-27) = 3\cdot(-27+3) = 3\cdot(-24) = -72$.
• For $x=-9$: $f(-9) = 0\cdot(-9+3) = 0$.
• For $x=21$: $f(21) = 5\cdot(21+3) = 5\cdot24 = 120$.

These give three points on the graph of $y=f(x)$: $(-27,-72)$, $(-9,0)$, and $(21,120)$. (They must be collinear because $f$ is linear.)

Use any two of the points to find the slope $m$.

Using $(-9,0)$ and $(21,120)$:

So the line $y=f(x)$ has slope $m=4$.

Write $f(x)$ in the form $f(x) = mx + b$.

We already know $m=4$, so $f(x) = 4x + b$. Use one of the points on the line, for example $(-9,0)$:

Thus $f(x) = 4x + 36$, and the $y$-intercept is the point where $x=0$, which is $(0, 36)$.