Use the fact that the horizontal asymptote is $y=5$ to write an equation involving $a$. What value of $a$ does this give you?

Once you know $a$, plug $x=3$ and $g(3)=1$ into $g(x)=\dfrac{ax+b}{x+2}$ to solve for $b$. Then think about whether $b$ is positive or negative.

After you find specific values for $a$ and $b$, compare their signs to I ($a>0$) and II ($b<0$). Decide which of these inequalities are always satisfied.

In Desmos, type y = (a x + b)/(x + 2); Desmos will create sliders for $a$ and $b$.

On a new line, type y = 5 to show the horizontal asymptote, and on another line type (3,1) to plot the point the graph must pass through.

Move the slider for $a$ until the ends of the graph of $y = (a x + b)/(x + 2)$ line up with the horizontal line $y=5$ as $x$ goes far left and far right. Note the value and sign of $a$ when this happens.

With $a$ fixed from the previous step, move the slider for $b$ until the curve goes exactly through the point $(3,1)$. Then observe whether $b$ is positive or negative and compare this with statements I and II.

For a rational function where the numerator and denominator are both linear (degree 1), like

the horizontal asymptote is the ratio of the leading coefficients.

• The leading coefficient of the numerator is $a$.
• The leading coefficient of the denominator is $1$ (from $x+2$).

So the horizontal asymptote is

We are told the horizontal asymptote is $y=5$, so

Now we know $a$ exactly.

The graph passes through $(3,1)$, which means that when $x=3$, $g(x)=1$.

Substitute $x=3$, $g(3)=1$, and $a=5$ into the function:

Simplify step by step:

Multiply both sides by $5$:

Subtract $15$ from both sides:

So $b = -10$.

We found:

• $a = 5$, which is greater than $0$, so statement I ($a>0$) is true.
• $b = -10$, which is less than $0$, so statement II ($b<0$) is also true.

Therefore, both I and II must be true, so the correct answer choice is C) I and II.