For each rewritten equation, think of it as an intersection of two graphs: $y=3^x$ with a horizontal line, with a straight line, or with a parabola. How many times can each pair meet?

For $q(x)$ and $r(x)$, plug in a few easy values (like $x=-1,0,1,2,3$) and look for where the function values change sign (from positive to negative or negative to positive). Each sign change across an interval guarantees at least one zero in that interval.

After you find where a function crosses once or twice, think about how it behaves for very large positive and very large negative $x$. Does the curve ever have a chance to come back and cross the axis again?

In Desmos, enter the three functions:

• y = 7 - 3^x
• y = 3^x - 6x
• y = x^2 - 3^x Make sure you can see where each graph meets the x-axis.

Look at the graph of y = 7 - 3^x and find how many times it touches or crosses the x-axis. You can tap the intercept point to see its exact x-coordinate, but you only need the number of intercepts.

Now look at the graph of y = 3^x - 6x. Observe how many distinct x-intercepts there are (where the curve crosses the x-axis). This tells you how many real zeros function II has.

Finally, examine the graph of y = x^2 - 3^x. Note how many times it crosses the x-axis and whether it ever comes back to cross again as you move left or right. Use this to decide how many real zeros function III has, then pick the answer choice that matches the counts from all three graphs.

A real zero of a function is an $x$-value where the function equals $0$ (an $x$-intercept).

Set each function equal to $0$:

• I. $7 - 3^x = 0$ ⟹ $3^x = 7$
• II. $3^x - 6x = 0$ ⟹ $3^x = 6x$
• III. $x^2 - 3^x = 0$ ⟹ $x^2 = 3^x$

We will use the shapes of the graphs $y=3^x$, $y=6x$, $y=x^2$, and horizontal lines to count how many real solutions each equation has.

Equation: $7 - 3^x = 0$ ⟹ $3^x = 7$.

Key facts about $3^x$:

• $3^x$ is always positive.
• $3^x$ is strictly increasing (as $x$ increases, $3^x$ increases without ever turning around).

So the graph of $y=3^x$ crosses the horizontal line $y=7$ exactly once. That means:

• $3^x = 7$ has exactly one real solution (you could call it $x = \log_3 7$).
• Therefore, $p(x)$ has exactly one real zero.

Equation: $3^x - 6x = 0$ ⟹ $3^x = 6x$.

Compare the functions $y=3^x$ (exponential) and $y=6x$ (line) by checking values of $q(x)$:

• $q(-1) = 3^{-1} - 6(-1) = \tfrac{1}{3} + 6 > 0$
• $q(0) = 1 - 0 = 1 > 0$
• $q(1) = 3 - 6 = -3 < 0$ ⟹ sign change between $0$ and $1$ → one zero in $(0,1)$
• $q(2) = 9 - 12 = -3 < 0$
• $q(3) = 27 - 18 = 9 > 0$ ⟹ sign change between $2$ and $3$ → another zero in $(2,3)$

So $q(x)$ is positive, then negative, then positive again, which forces it to cross the x-axis at least twice.

Graphically, $y=3^x$ is a curve that bends upward and gets steeper and steeper, while $y=6x$ is a straight line with constant slope. Such a curve and a line can intersect at most twice, so $q(x)$ has exactly two real zeros, not one.

Equation: $x^2 - 3^x = 0$ ⟹ $x^2 = 3^x$.

First, look for a sign change to guarantee a zero:

• $r(-1) = (-1)^2 - 3^{-1} = 1 - \tfrac{1}{3} = \tfrac{2}{3} > 0$
• $r(0) = 0^2 - 1 = -1 < 0$

Because $r(x)$ is continuous, this sign change between $x=-1$ and $x=0$ means there is one real zero in $(-1,0)$.

Now check behavior to the left and right so we can see if there are any more zeros:

• For large negative $x$, $x^2$ is very large positive while $3^x$ is very close to $0$, so $r(x) = x^2 - 3^x$ is large positive and stays positive as you move further left. There is no extra crossing on the far left.

• For $x \ge 0$, compare growth:

  • At $x=0$: $r(0)=-1$.
  • At $x=1$: $r(1)=1-3=-2$.
  • At $x=2$: $r(2)=4-9=-5$.
  • At $x=3$: $r(3)=9-27=-18$.

  As $x$ increases, $3^x$ multiplies by $3$ each step, while $x^2$ grows much more slowly, so $3^x$ pulls farther and farther ahead of $x^2$. That means $x^2 - 3^x$ becomes more and more negative and never comes back up to $0$.

So $r(x)$ crosses the x-axis once between $-1$ and $0$ and then stays negative for all larger $x$, giving exactly one real zero overall.

Summary:

• I: $7 - 3^x$ → exactly one real zero.
• II: $3^x - 6x$ → two real zeros.
• III: $x^2 - 3^x$ → exactly one real zero.

So the functions with exactly one real zero are I and III, which corresponds to answer choice C) I and III.