Once you write $9$ as $3^2$, you will have $3^{x-2} = 3^2$. What can you say about the exponents when the bases are the same?

After you solve for $x$ from the exponential equation, substitute that $x$ into $y = x^2 - 6x + 13$ and simplify carefully.

When you plug in $x$, compute $x^2$, then $6x$, and combine the numbers step by step to avoid sign errors.

In Desmos, enter y = 3^(x-2) and y = 9. Look for the point where these two graphs intersect and note the x-coordinate of that intersection; that is the value of $x$ that satisfies $3^{x-2} = 9$.

In a new line, type y = x^2 - 6x + 13. Then either (a) click on the intersection point from step 1 and read off its x-coordinate, and substitute that value into x^2 - 6x + 13 in a separate expression, or (b) create a table for y = x^2 - 6x + 13 and enter that x-value in the table. The corresponding output is the $y$-value that answers the question.

The system

represents all $(x,y)$ that satisfy both equations at the same time. To find $y$, we first need to find the corresponding $x$ from the second equation and then plug that $x$ into the first equation.

Start with the second equation:

Rewrite $9$ as a power of $3$:

So the equation becomes

Since the bases are the same and both sides are positive, set the exponents equal:

Solve for $x$:

This is the $x$-coordinate of the solution to the system.

Now use the first equation, $y = x^2 - 6x + 13$, and substitute $x = 4$:

Compute each part step by step:

• $4^2 = 16$
• $6(4) = 24$ So

Combine the terms carefully:

So the solution to the system has $y$-value

Therefore, the correct answer is $5$.