Question

complete the table to investigate dilations of exponential functions.

$x$	$2^x$	$3\cdot 2^x$	$2^{3x}$
$-1$	$\frac{1}{2}$	$\frac{3}{2}$	$\frac{1}{8}$
$0$	$a$	$b$	$c$
$1$	$d$	$e$	$f$
$2$	$4$	$12$	$64$

$a=\square$ $b=\square$
$c=\square$ $d=\square$
$e=\square$ $f=\square$

Explanation:

Step1: Find \(a\) (when \(x = 0\) for \(2^x\))

For the function \(y = 2^x\), substitute \(x = 0\). Using the property of exponents \(a^0=1\) (where \(a
eq0\)), we have \(2^0 = 1\). So \(a = 1\).

Step2: Find \(b\) (when \(x = 0\) for \(3\cdot2^x\))

For the function \(y = 3\cdot2^x\), substitute \(x = 0\). We know \(2^0 = 1\), so \(3\cdot2^0=3\times1 = 3\). Thus \(b = 3\).

Step3: Find \(c\) (when \(x = 0\) for \(2^{3x}\))

For the function \(y = 2^{3x}\), substitute \(x = 0\). Then \(2^{3\times0}=2^0 = 1\). So \(c = 1\).

Step4: Find \(d\) (when \(x = 1\) for \(2^x\))

For the function \(y = 2^x\), substitute \(x = 1\). We get \(2^1 = 2\). So \(d = 2\).

Step5: Find \(e\) (when \(x = 1\) for \(3\cdot2^x\))

For the function \(y = 3\cdot2^x\), substitute \(x = 1\). We know \(2^1 = 2\), so \(3\cdot2^1=3\times2 = 6\). Thus \(e = 6\).

Step6: Find \(f\) (when \(x = 1\) for \(2^{3x}\))

For the function \(y = 2^{3x}\), substitute \(x = 1\). Then \(2^{3\times1}=2^3 = 8\). So \(f = 8\).

Answer:

\(a = 1\), \(b = 3\), \(c = 1\), \(d = 2\), \(e = 6\), \(f = 8\)