Exponential Functions and Graphs: Additional Practice (Set 1)

The inputs are three apart, so dividing the outputs isolates $b^{3}$ and cancels $a$.

$b^{3} = \frac{96}{12} = 8 \quad \Longrightarrow \quad b = 2$

Now use $f(1) = ab = 12$, so $2a = 12$ and $a = 6$. Since $f(0) = a \cdot b^{0} = a$, the answer is $f(0) = 6$.

Rewrite it as $y = -2^{x} + 5$ so the transformations are visible: the parent $2^{x}$ is flipped by the minus sign and then shifted up $5$.

$2^{x} \text{ increases} \quad \Longrightarrow \quad -2^{x} \text{ decreases}$

As $x$ runs off to the left, $2^{x}$ shrinks toward $0$ and $y$ rises toward $5$ without reaching it, so the asymptote is $y = 5$ and the curve falls from left to right.

The negative exponent flips the fraction, and then the denominator $3$ is a cube root and the numerator $2$ is a square.

$\left(\frac{8}{27}\right)^{-2/3} = \left(\frac{27}{8}\right)^{2/3} = \left(\sqrt[3]{\frac{27}{8}}\right)^{2} = \left(\frac{3}{2}\right)^{2} = \frac{9}{4}$

The result is positive, as every value of an exponential must be. The distractor $\tfrac49$ is what you get by forgetting to flip.

Reflecting across the $y$-axis replaces the input $x$ with $-x$.

$y = 3^{-x} = \left(3^{-1}\right)^{x} = \left(\tfrac{1}{3}\right)^{x}$

So the growth curve with base $3$ becomes the decay curve with base $\tfrac13$, which is why every decaying exponential is a growing one seen in a mirror. Reflecting across the $x$-axis would instead give $y = -3^{x}$.

Split the exponent, then factor out the common power.

$2^{\,x+3} - 2^{x} = 2^{x} \cdot 2^{3} - 2^{x} = 2^{x}(8 - 1) = 7 \cdot 2^{x}$

The difference is not a constant; it is itself proportional to $2^{x}$, which is why differences are the wrong tool for exponentials and ratios are the right one. (The option $2^{3} \cdot 2^{x}$ is $f(x+3)$ alone, with the subtraction forgotten.)

A shift of $h$ units right replaces $x$ with $x - h$ inside the function.

$y = 2^{\,x-4}$

Check a point: the parent passes through $(0, 1)$, and the shifted curve should pass through $(4, 1)$. Indeed $2^{\,4-4} = 2^{0} = 1$. Notice this is also a vertical squash, since $2^{\,x-4} = \tfrac{1}{16} \cdot 2^{x}$.

A quotient of like powers is the power of the quotient.

$\frac{6^{x}}{2^{x}} = \left(\frac{6}{2}\right)^{x} = 3^{x}$

The bases divide but the exponent stays put. Check at $x = 2$: $\tfrac{36}{4} = 9 = 3^{2}$. Subtracting the bases to get $4^{x}$ is the classic error.

Substitute the point into the rule.

$b^{3} = \frac{1}{8} \quad \Longrightarrow \quad b = \sqrt[3]{\frac{1}{8}} = \frac{1}{2}$

The base is between $0$ and $1$, matching the fact that the output at a positive input has fallen below $1$. A base can never be negative, so $-\tfrac12$ is not admissible (and its cube is $-\tfrac18$ anyway).

Write the right side over the base $2$.

$8^{\,x-1} = \left(2^{3}\right)^{x-1} = 2^{\,3x-3}$

One-to-oneness lets you equate the exponents, which leaves a quadratic.

$x^{2} - 1 = 3x - 3 \quad \Longrightarrow \quad x^{2} - 3x + 2 = 0 \quad \Longrightarrow \quad (x-1)(x-2) = 0$

So $x = 1$ or $x = 2$. Both check out: at $x = 1$ both sides are $2^{0} = 1$, and at $x = 2$ both sides are $2^{3} = 8$.

Form the ratio of consecutive outputs and watch the coefficient cancel.

$\frac{f(x+1)}{f(x)} = \frac{7 \cdot \left(\tfrac35\right)^{x+1}}{7 \cdot \left(\tfrac35\right)^{x}} = \frac{3}{5}$

The factor is the base $\tfrac35$, no matter where you start. Since $\tfrac35 < 1$, each step shrinks the output, so this function decays.

Doubling the input doubles the exponent, and a doubled exponent is a squared value.

$g(2x) = 3^{2x} = \left(3^{x}\right)^{2} = \left[g(x)\right]^{2}$

Check at $x = 2$: $g(4) = 81$ and $\left[g(2)\right]^{2} = 9^{2} = 81$. The option $9g(x)$ would be $g(x+2)$, which adds to the input rather than doubling it.

Look for inputs where the two rules agree.

$x = 2:\; 2^{2} = 4 = 2^{2}, \qquad x = 4:\; 2^{4} = 16 = 4^{2}$

Both are genuine crossings. Between them the parabola is ahead (at $x = 3$, $9 > 8$), and outside them the exponential is ahead (at $x = 1$, $2 > 1$; at $x = 5$, $32 > 25$). Past $x = 4$ the exponential doubles at each step while the square grows by a shrinking factor, so it never gives the lead back and there is no third crossing.