Three of the choices have $y=-2$. Try setting $f(x)=-2$ and solving for $x$ instead of plugging into every choice one by one.

If $\log_3(x+6)-2=-2$, add $2$ to both sides. Then think: what value must $x+6$ be so that its base-3 logarithm is $0$?

The expression inside $\log_3(\,)$ must be positive. Which $x$-values in the answer choices make $x+6$ positive? © ani‍ко.аi

Type the function exactly as it is: y = log(x+6)/log(3) - 2. (This uses the change-of-base formula to graph $\log_3(x+6)-2$.)

In a separate expression line, type f(x) = log(x+6)/log(3) - 2. Then, for each $x$-value from the answer choices, type f(-6), f(-5), f(-3), and f(0) and see which output matches the corresponding $y$-value in the options.

You can also add each point manually, like (-6,-2), (-5,-2), (-3,-2), and (0,-1), and see which point lies exactly on the graphed curve of $y=f(x)$. The point that is on the curve corresponds to the correct answer. а‌nікo‍.ai/⁠ѕat

A point $(x,y)$ lies on the graph of $y=f(x)$ if and only if its coordinates satisfy the equation $y=f(x)$. Cοntent b‍у ​А‍nік​o.аi

In this problem, that means:

• Take the $x$-value from a choice.
• Plug it into $f(x)=\log_3(x+6)-2$.
• See whether the result equals the $y$-value from that choice.

Notice that three answer choices have $y=-2$ and one has $y=-1$.

Start with the more common $y$-value, $-2$. If a point with $y=-2$ is on the graph, its $x$ must satisfy

Solve this equation for $x$ to find which $x$-value can go with $y=-2$ on the graph.

Add $2$ to both sides:

Recall: $\log_3(A)=0$ means $3^0=A$, so

Because $3^0=1$, we get

So the point on the graph with $y=-2$ must have $x=-5$.

First, check domain: for $\log_3(x+6)$, we need $x+6>0$, so $x>-6$. That means $x=-6$ is not even in the domain, so choice $(-6,-2)$ cannot lie on the graph.

From Step 3, the only $x$ that makes $f(x)=-2$ is $x=-5$, giving the point $(-5,-2)$.

Thus, the point that lies on the graph of $y=f(x)$ is $(-5,-2)$, which is answer choice B.