Look at the three equations you formed. How can you combine them (for example, by subtracting one from another) so that $c$ disappears?

After eliminating $c$, you will have two equations in $a$ and $b$. Try dividing one by the other to cancel $a$ and solve for $b$ using the given condition $b>0$ and $b\neq 1$.

Remember that $b^0=1$. Once you know $a$, $b$, and $c$, write $g(0)=ab^0+c$ in its simplest form and evaluate it.

Create a table with the three points: in the first column (x1), enter -1, 1, and 3; in the second column (y1), enter 10, 40, and 160.

In a new line, type y1 ~ A*B^(x1) + C (using capital letters for the parameters). Desmos will perform an exponential regression and display numerical values for $A$, $B$, and $C$ that model $g(x)=A B^x + C$.

In another line, type A*B^0 + C. The value Desmos shows for this expression is the value of $g(0)$ according to the fitted function; read this number carefully as your answer.

From $g(x)=ab^x+c$ and the table:

• For $x=-1$: $g(-1)=10$ gives $ab^{-1}+c=10$, so $\dfrac{a}{b}+c=10$.
• For $x=1$: $g(1)=40$ gives $ab+c=40$.
• For $x=3$: $g(3)=160$ gives $ab^3+c=160$.

Now you have a system of three equations in $a$, $b$, and $c$.

Subtract equations so that $c$ cancels out.

• Subtract the $x=-1$ equation from the $x=1$ equation:

• Subtract the $x=1$ equation from the $x=3$ equation:

Now you have:

• $a\left(b - \dfrac{1}{b}\right) = 30$
• $a(b^3 - b) = 120$.

Divide the second equation by the first to eliminate $a$:

The $a$ cancels:

Simplify the fraction:

• $b^3 - b = b(b^2 -1)$
• $b - \dfrac{1}{b} = \dfrac{b^2 -1}{b}$

So

Thus $b^2 = 4$, so $b=2$ or $b=-2$. Since the problem states $b>0$, we must have $b=2$.

Use $b=2$ in one of the simpler equations, for example

Substitute $b=2$:

Now plug $a=20$ and $b=2$ into $ab+c=40$:

So $g(x)=20\cdot 2^x + 0 = 20\cdot 2^x$.

Use $g(x)=ab^x+c$ with $a=20$, $b=2$, and $c=0$.

So the value of $g(0)$ is $20$, which corresponds to choice B.