Exponential Functions and Graphs: Free Response

Divide each output by the one before it.

$\frac{88}{11} = 8, \qquad \frac{704}{88} = 8, \qquad \frac{5632}{704} = 8.$

All three ratios come out to the same number, $8$.

A step of size $s$ multiplies the output by $b^{s}$, not by $b$ itself. Every step in this table is $3$ units, so the constant ratio $8$ found in part A is $b^{3}$, not $b$:

$b^{3} = 8 \quad \Longrightarrow \quad b = 2.$

The output at $x=0$ is the multiplier $a$ directly, since $f(0) = a\cdot b^{0} = a$. Reading the table, $a = 11$. The rule is

$f(x) = 11\cdot 2^{x}.$

Check: $f(3) = 11\cdot 8 = 88$ and $f(9) = 11\cdot 512 = 5632$, both matching the table.

The classmate mistook the three-step ratio for the base itself, but a ratio only equals the base when the inputs are exactly one unit apart. Here the spacing is $3$, so the ratio $8$ is $b^{3}$, and the true base recovered in part B is $b=2$.

Sampling the same function $f(x) = 11\cdot 2^{x}$ at consecutive integers instead, the ratio between neighbours would be exactly the base:

$\frac{f(x+1)}{f(x)} = b = 2.$

So unit-spaced sampling would have shown a constant ratio of $2$, not $8$.

Divide the larger value by the smaller. The unknown $a$ appears in both and cancels:

$\frac{f(6)}{f(2)} = \frac{a\cdot b^{6}}{a\cdot b^{2}} = b^{4} = \frac{5103}{63} = 81.$

A base is positive, so $b$ is the positive fourth root of $81$:

$b = 3.$

Substitute back into $f(2)=63$:

$a\cdot 3^{2} = 63 \quad\Longrightarrow\quad 9a = 63 \quad\Longrightarrow\quad a = 7.$

So $f(x) = 7\cdot 3^{x}$. Check: $f(2)=7\cdot9=63$ and $f(6)=7\cdot729=5103$.

Set the rule equal to the target value and isolate the power of $3$:

$7\cdot 3^{x} = 567 \quad\Longrightarrow\quad 3^{x} = 81.$

Write $81$ as a power of the same base, $81 = 3^{4}$, so the equation reads $3^{x} = 3^{4}$. Since an exponential function is one-to-one, the exponents must match:

$x = 4.$

In $f(x) = a\cdot b^{x}$, the constant $a$ multiplies every output by the same amount, so when one output is divided by another, that shared factor cancels:

$\frac{a\cdot b^{6}}{a\cdot b^{2}} = b^{4}.$

If both readings had instead come from the very same input, the division would compare $f(2)$ to $f(2)$, giving

$\frac{f(2)}{f(2)} = 1 = b^{0},$

which holds for every possible base $b$. The two readings would carry no information about which base the function actually has, because a genuine constraint on $b$ only appears when the two inputs are different.

Directly, $f(1) = -9$ and $f(2) = (-9)^{2} = 81$.

The law of exponents forces a relationship between $f\!\left(\tfrac12\right)$ and $f(1)$: splitting the input in half,

$f(1) = f\!\left(\tfrac12+\tfrac12\right) = f\!\left(\tfrac12\right)\cdot f\!\left(\tfrac12\right) = \left[f\!\left(\tfrac12\right)\right]^{2}.$

So $\left[f\!\left(\tfrac12\right)\right]^{2}$ would have to equal $f(1) = -9$. A square of a real number is never negative, so no real number can play the role of $f\!\left(\tfrac12\right)$. The function simply has no value there.

Stack four copies of the three-quarter step to reach $3$: $\tfrac34+\tfrac34+\tfrac34+\tfrac34=3$, so the law of exponents gives

$f(3) = \left[f\!\left(\tfrac34\right)\right]^{4}.$

Here $f(3) = (-9)^{3} = -729$, so $\left[f\!\left(\tfrac34\right)\right]^{4}$ would have to equal $-729$. A fourth power of a real number is never negative, so this input fails exactly the same way $x=\tfrac12$ did.

An exponential function, by definition, must be defined for every real input, not merely most of them. Finding a single real number where no value can be assigned is already enough to disqualify $b=-9$, regardless of how ordinary the function looks at whole numbers.

For any real $x$ and $y$, $1^{x+y}=1$, and also $1^{x}\cdot 1^{y} = 1\cdot 1 = 1$, so

$1^{x+y} = 1^{x}\cdot 1^{y}$

holds for every choice of $x$ and $y$ whatsoever. The law of exponents is never violated.

Every genuine exponential function is one-to-one: $b^{u}=b^{v}$ forces $u=v$, because the function is strictly increasing or strictly decreasing. The base $b=1$ breaks exactly this. Since $f(x)=1^{x}=1$ for every $x$, the equation $1^{u}=1^{v}$ is true for every pair $u,v$, not only equal ones, so $f$ takes the same output at every input and equating exponents tells you nothing. That is why $b=1$ is excluded: not because the law of exponents fails, but because the resulting function is flat and carries none of the behaviour the rest of the theory relies on.

As $x$ runs off to the left, $2^{\,x+2}$ shrinks toward $0$ while staying positive, so

$g(x) = -5\cdot 2^{\,x+2} - 4 \;\longrightarrow\; -5\cdot 0 - 4 = -4.$

The horizontal asymptote is $y=-4$.

Since $2^{\,x+2}$ is always strictly positive, $-5\cdot 2^{\,x+2}$ is always strictly negative, so $g(x) = -4 + (\text{a negative number})$ stays below $-4$ for every $x$. The range is all $y < -4$.

The $y$-intercept comes from evaluating at $x=0$:

$g(0) = -5\cdot 2^{2} - 4 = -5\cdot4-4 = -24.$

An $x$-intercept would require $g(x)=0$ for some $x$, but part A showed every output of $g$ satisfies $y<-4$. Since $0$ is not less than $-4$, no input can produce an output of $0$, so the graph never crosses the $x$-axis.

The base is $2$, which is greater than $1$, so the parent expression $2^{\,x+2}$ is strictly increasing in $x$. Multiplying a strictly increasing expression by a negative constant reverses that direction:

$2^{\,x+2}\ \text{increasing} \quad\Longrightarrow\quad -5\cdot 2^{\,x+2}\ \text{decreasing}.$

Subtracting the further constant $4$ does not affect direction at all, since it shifts every output by the same fixed amount. So $g$ is strictly decreasing.

Split the exponent using the law of exponents:

$2^{\,x+2} = 2^{x}\cdot 2^{2} = 4\cdot 2^{x}.$

Substituting back,

$g(x) = -5\cdot\left(4\cdot 2^{x}\right) - 4 = -20\cdot 2^{x} - 4.$

So shifting $-5\cdot 2^{x}-4$ two units to the left produces exactly the same graph as multiplying only its exponential part by $4$, leaving the vertical shift of $-4$ untouched: the displacement from the horizontal asymptote $y=-4$ is stretched by a factor of $4$. This matches the general fact that a horizontal shift by $h$ in $b^{\,x-h}$ is the same as multiplying by the constant $b^{-h}$; here $h=-2$ and $2^{-(-2)}=2^{2}=4$.

$f(8)=256,\ f(9)=512,\ f(10)=1024,\ f(11)=2048.$

$p(8)=512,\ p(9)=729,\ p(10)=1000,\ p(11)=1331.$

Comparing pairwise: at $x=8$, $256<512$, so $p$ leads. At $x=9$, $512<729$, so $p$ still leads. At $x=10$, $1024>1000$, so $f$ has taken over. At $x=11$, $2048>1331$, and $f$ leads by even more.

The per-step factor for $p$ from $x$ to $x+1$ is $\left(\dfrac{x+1}{x}\right)^{3}$. At $x=9$:

$\left(\frac{10}{9}\right)^{3} = \frac{1000}{729} \approx 1.372.$

At $x=99$:

$\left(\frac{100}{99}\right)^{3} = \frac{1000000}{970299} \approx 1.031.$

Both are already smaller than $f$'s per-step factor, the fixed base $2$, and this per-step factor for $p$ only keeps shrinking toward $1$ as $x$ grows further, since $\dfrac1x$ shrinks. So from $x=9$ onward, every single step multiplies $f$ by exactly $2$ while multiplying $p$ by something smaller than $2$, and getting smaller still. Once $f$ takes the lead at $x=10$, each further step widens the gap rather than closing it, so the lead can never pass back to $p$.

For a polynomial of any degree $n$, the ratio of consecutive values $\dfrac{p(x+1)}{p(x)}$ tends toward $1$ as $x$ grows, because the extra unit added to $x$ matters less and less compared to $x$ itself once $x$ is large. A higher degree only postpones this, by letting the ratio start out large. An exponential function with base $b>1$, in contrast, multiplies by the exact same factor $b$ at every step, forever, with no shrinking at all:

$\frac{p(x+1)}{p(x)} \longrightarrow 1 \quad \text{as } x \to \infty, \qquad \frac{f(x+1)}{f(x)} = b \ \text{(fixed, for every } x\text{)}.$

Since a fixed number greater than $1$ and a quantity shrinking toward $1$ must eventually cross, there is always some point past which the exponential's fixed step beats the polynomial's step, and every step after that widens the gap further rather than closing it. This is exactly why no polynomial, of any degree, can keep pace with a genuine exponential function forever.