Notice that when $x$ increases by 2, the function value is multiplied by the same number. What should happen to $f(x)$ when $x$ goes from 4 to 6?

For $f(x)=ae^{bx}+c$, use the equations from $x=0,2,4$ to eliminate $c$ and find $e^{2b}$. Then use that to predict $f(6)$ without fully solving for all constants if you prefer.

In Desmos, first verify the growth factor: type 27/9 and 81/27; you should see they are the same number. Then, to extend the pattern from $x=4$ to $x=6$, type 81 * 3 into Desmos and use the value that appears as the function value at $x=6$.

To confirm the algebraic model, you can type 9*e^((ln(3)/2)*6) into Desmos (this corresponds to $f(6)=9e^{(\ln 3/2)\cdot 6}$). The output of this expression is the value of $f(6)$.

You are given three values: $f(0)=9$, $f(2)=27$, and $f(4)=81$.

Check how the outputs change as $x$ increases by 2:

• From $x=0$ to $x=2$: $27/9=3$.
• From $x=2$ to $x=4$: $81/27=3$.

So, every time $x$ increases by 2, the value of $f(x)$ is multiplied by 3. This is the hallmark of an exponential pattern.

From $x=4$ to $x=6$, $x$ again increases by 2. Based on the pattern you found, $f(x)$ should be multiplied by 3 once more.

So starting from $f(4)=81$ and increasing $x$ by 2, compute $f(6)$ by multiplying 81 by 3.

To confirm, assume $f(x)=ae^{bx}+c$ fits the pattern.

1. At $x=0$: $f(0)=ae^{0}+c=a+c=9$.
2. At $x=2$: $f(2)=ae^{2b}+c=27$.
3. At $x=4$: $f(4)=ae^{4b}+c=81$.

Subtract $f(0)$ from $f(2)$ and $f(4)$ to eliminate $c$:

• $f(2)-f(0)=ae^{2b}-a=a(e^{2b}-1)=27-9=18$.
• $f(4)-f(2)=ae^{4b}-ae^{2b}=ae^{2b}(e^{2b}-1)=81-27=54$.

Divide the second equation by the first:

Then $2b=\ln 3$ and $b=\tfrac{1}{2}\ln 3$. From $a+c=9$ and $ae^{2b}+c=27$ with $e^{2b}=3$:

• $ae^{2b}=3a=27-c$,
• $a+c=9$. Solving gives $c=0$ and $a=9$, so $f(x)=9e^{(\ln 3/2)x}$, which matches the pattern you used.

Now use the pattern from Step 2: $f(6)=81\times 3=243$, so the correct answer is 243.