Exponential Functions and Graphs: Practice

The point at $x = 0$ hands you the coefficient directly, since $b^{0} = 1$, so $a = 7$.

A one-unit step multiplies the output by the base, so divide the two outputs.

$b = \frac{f(1)}{f(0)} = \frac{21}{7} = 3$

The rule is $f(x) = 7 \cdot 3^{x}$.

A step of size $s$ multiplies the output by $b^{s}$, and here the step from $2$ to $4$ has size $s = 2$.

$\frac{f(4)}{f(2)} = \frac{2 \cdot 3^{4}}{2 \cdot 3^{2}} = 3^{2} = 9$

As a check, $f(4) = 162$ and $f(2) = 18$, and $162 \div 18 = 9$. The coefficient $2$ always cancels.

Replacing the input $x$ by $-x$ reflects a graph across the $y$-axis, and that is exactly what happened here.

$5^{-x} = \left(5^{-1}\right)^{x} = \left(\tfrac{1}{5}\right)^{x}$

So the growth curve with base $5$ becomes the decay curve with base $\tfrac15$. A reflection across the $x$-axis would instead give $y = -5^{x}$, which is negative everywhere.

The parent value $2^{x}$ is strictly positive for every real $x$, never zero.

$2^{x} > 0 \quad \Longrightarrow \quad 2^{x} + 5 > 5$

So the outputs are everything above $5$ but not $5$ itself, since the asymptote $y = 5$ is approached and never reached.

Neither side is a power of the other, but both are powers of $2$.

$4^{x} = \left(2^{2}\right)^{x} = 2^{2x}, \qquad 8 = 2^{3}$

So $2^{2x} = 2^{3}$, and one-to-oneness forces $2x = 3$, giving $x = \tfrac32$. Check: $4^{3/2} = \left(\sqrt{4}\right)^{3} = 8$.

Divide the two values so that $a$ cancels and only a power of $b$ is left.

$\frac{f(3)}{f(1)} = b^{2} = \frac{54}{6} = 9 \quad \Longrightarrow \quad b = 3$

Now substitute back into $f(1) = ab = 6$, which gives $3a = 6$ and $a = 2$. The rule is $f(x) = 2 \cdot 3^{x}$.

Substitute the point into the rule.

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

Bases are positive, so the positive cube root is the one to take. The curve is $y = 2^{x}$, and $2^{3} = 8$ confirms it.

Split the exponent with the law of exponents and evaluate the constant piece.

$2^{\,x-3} = 2^{x} \cdot 2^{-3} = \frac{1}{8} \cdot 2^{x}$

This is the surprise that belongs to exponentials alone: shifting the graph three units right is the same as squashing it to one eighth of its height. Subtracting in the exponent never subtracts from the output.

An exponential function has the form $a \cdot b^{x}$ with the variable in the exponent and a fixed positive base other than $1$.

$f(x) = x^{2} \quad \text{has the variable in the } \textbf{base}$

That makes it a power function (a parabola). The other three all put the variable in the exponent, with bases $2$, $5$, and $\tfrac13$.

Evaluate at $x = 0$, remembering that any base to the power $0$ is $1$.

$g(0) = 4 \cdot 3^{0} - 5 = 4 \cdot 1 - 5 = -1$

The intercept is $(0, -1)$. The value $-5$ is the horizontal asymptote, which the curve approaches on the far left but never attains.

Split the exponent with the law of exponents.

$f(x+2) = 3^{\,x+2} = 3^{x} \cdot 3^{2} = 9 \cdot 3^{x} = 9 f(x)$

Adding to the input multiplies the output; it never adds to the output. Stepping two units right always multiplies by $b^{2} = 9$, whatever the starting point.

Every real input has a value. Whole-number inputs come from repeated multiplication, negative inputs from reciprocals, and fractions from roots.

$2^{-3} = \tfrac18, \qquad 2^{0} = 1, \qquad 2^{7/5} = \sqrt[5]{2^{7}}$

An irrational input such as $\sqrt{2}$ is squeezed between rational exponents ($2^{1.414} < 2^{\sqrt 2} < 2^{1.415}$), which pins down a single value. So the domain is all of the real numbers.