Exponential Functions and Graphs: Practice

An exponential function has the variable in the exponent and a fixed number as the base.

$f(x) = 5^{x} \quad \text{(variable exponent, fixed base } 5)$

In $x^{5}$ and $x^{1/5}$ the variable is the base and the exponent is fixed, so those are power functions, and $5x$ is linear.

Substitute $x = 4$ into the exponent and multiply four copies of the base.

$3^{4} = 3 \cdot 3 \cdot 3 \cdot 3 = 81$

The distractor $12$ comes from computing $3 \cdot 4$, which would be the linear function, not the exponential one.

The law of exponents forces the value at $0$. Split $1$ as $1 + 0$.

$7 = 7^{1} = 7^{1+0} = 7^{1} \cdot 7^{0} = 7 \cdot 7^{0}$

Dividing both sides by $7$ gives $7^{0} = 1$. Every exponential passes through $(0, 1)$, whatever its base.

The exponent $\tfrac12$ asks for the square root, and an exponential is always positive, so take the positive root.

$9^{1/2} = \sqrt{9} = 3$

Check it against the law of exponents: $9^{1/2} \cdot 9^{1/2} = 9^{1} = 9$, and indeed $3 \cdot 3 = 9$.

Run the input off to the left and watch the output shrink.

$2^{-1} = \tfrac12, \qquad 2^{-10} = \tfrac{1}{1024}, \qquad 2^{-100} = \tfrac{1}{2^{100}}$

The values get as close to $0$ as you like but stay positive, so the curve flattens toward the line $y = 0$ without ever touching it.

Every value of an exponential is a square, since halving the input and applying the law of exponents gives

$4^{x} = \left(4^{x/2}\right)^{2} > 0.$

So the output is never negative and never zero, while the unbroken curve climbs from values near $0$ to values as large as you like. The range is every positive real number.

Evaluate the power first, then multiply by the coefficient.

$f(0) = 2 \cdot 3^{0} = 2 \cdot 1 = 2$

In $f(x) = a \cdot b^{x}$ the coefficient $a$ is always the starting value, because $b^{0} = 1$ wipes out the base.

Write the base as a power of $2$ and use the rule for a power of a power.

$\left(\tfrac{1}{2}\right)^{x} = \left(2^{-1}\right)^{x} = 2^{-x}$

So halving repeatedly is the same as doubling backwards, which is why the two graphs are mirror images across the $y$-axis.

Each hour multiplies the count by the same factor $3$, so the growth is exponential with base $3$, and the starting count $4$ is the coefficient.

$P(t) = 4 \cdot 3^{t}$

Check the first two hours: $P(0) = 4$, $P(1) = 12$, $P(2) = 36$, each triple the one before. The option $4 + 3t$ adds $3$ each hour instead of multiplying.

Write out both values and cancel.

$\frac{f(x+1)}{f(x)} = \frac{a \cdot b^{x+1}}{a \cdot b^{x}} = b$

The $x$ vanishes completely, so a one-unit step multiplies the output by $b$ no matter where you start. That constant ratio is what makes the function exponential.

Divide each output by the one before it and look for a constant quotient.

$\frac{3}{1} = 3, \qquad \frac{9}{3} = 3, \qquad \frac{27}{9} = 3$

Only the first list has a constant ratio. The lists $1, 3, 5, 7$ and $3, 6, 9, 12$ have constant differences, so they are linear, and $1, 4, 9, 16$ has ratios $4$, $2.25$, $1.78\ldots$, which are the squares.

An exponent of $\tfrac12$ asks for a square root, and squaring a real number can never give a negative result.

$(-4)^{1/2} = \sqrt{-4} \quad \text{is not a real number}$

The other inputs are fine: $(-4)^{2} = 16$, $(-4)^{3} = -64$, and $(-4)^{0} = 1$. A negative base leaves a hole at every fraction with an even denominator, which is why bases must be positive.