Exponential Functions

$v(x) = 9{,}000(0.9)^x$

The starting value (at $x = 0$) is 9,000. Each value divided by the one before gives $8{,}100 \div 9{,}000 = 0.9$, so $v(x) = 9{,}000(0.9)^x$.

$v(x) = 9{,}000(1.1)^x$: The division was flipped. Divide each value by the previous one: 0.9, a decay factor.

$v(x) = 8{,}100(1.1)^x$: The starting value is at $x = 0$, not $x = 1$, and the factor is 0.9.

$v(x) = 8{,}100(0.9)^x$: The factor is right, but the starting value is the value at $x = 0$: 9,000.

$f(x) = 18\left(\frac{6}{5}\right)^x$

The initial value is $f(0) = 18$. Divide each value by the one before: $15 \div 12.5 = \frac{6}{5}$, and so on. So $f(x) = 18\left(\frac{6}{5}\right)^x$.

$f(x) = 15\left(\frac{5}{6}\right)^x$: The initial value is $f(0)$, not $f(-1)$, and the division was flipped.

$f(x) = 15\left(\frac{6}{5}\right)^x$: The factor is right, but the initial value is $f(0) = 18$.

$f(x) = 18\left(\frac{5}{6}\right)^x$: The values increase, so the factor must be greater than 1: $\frac{6}{5}$.

$g(x) = 6\left(\frac{1}{3}\right)^x$

The y-intercept is 6, so $a = 6$. Each y-value is $\frac{1}{3}$ of the one before ($6 \div 18$, $2 \div 6$), so $b = \frac{1}{3}$.

$g(x) = 6(3)^x$: The y-values were divided in the wrong order. Divide each value by the previous one: $\frac{1}{3}$.

$g(x) = 6 - \left(\frac{1}{3}\right)^x$: An exponential function multiplies: $a \cdot b^x$, not $a - b^x$.

$g(x) = 6 - (3)^x$: An exponential function has the form $a \cdot b^x$, and the factor here is $\frac{1}{3}$.

$p(x) = (0.25)^x$

The y-intercept is 1, so $a=1$. The next values are multiplied by 0.25, giving $p(x)=(0.25)^x$.

$p(x) = 2(0.5)^x$: This has y-intercept 2 and common factor 0.5; the graph has y-intercept 1 and factor 0.25.

$p(x) = (1.25)^x$: A base greater than 1 models growth, while this graph decays.

$p(x) = (25)^x$: A base of 25 produces rapid growth, not the decay shown.

$g(x)=25.5(0.6)^x$

At $x=0$, sales are 25.5 million, so $a=25.5$. The ratio $15.3/25.5=0.6$ gives the decay factor.

$g(x)=0.6(25.5)^x$: The initial value and decay factor are reversed.

$g(x)=6.12(25.5)^x$: The initial value is 25.5 at $x=0$, and the values decrease rather than grow.

$g(x)=25.5(6.12)^x$: The initial value is right, but the common factor is $15.3/25.5=0.6$.

$f(x)=16\left(\frac{3}{2}\right)^x$

Each value is multiplied by $\frac{54}{36}=\frac{3}{2}$. Since $f(2)=36$, $a(3/2)^2=36$, so $a=16$.

$f(x)=16\left(\frac{2}{3}\right)^x$: The table increases, so the base must be greater than 1.

$f(x)=36\left(\frac{3}{2}\right)^x$: 36 is $f(2)$, not the initial value $f(0)$.

$f(x)=36\left(\frac{2}{3}\right)^x$: The initial value is 16 and the values grow by factor $\frac{3}{2}$.

$P(x) = 90\left(\frac{1}{3}\right)^x$

In $P(x) = ab^x$, $a$ is the value when $x = 0$, so $a = 90$. Each value divided by the previous value is $\frac{90}{270} = \frac{30}{90} = \frac{10}{30} = \frac{1}{3}$, so $b = \frac{1}{3}$.

$P(x) = 30\left(\frac{1}{3}\right)^x$: The value of $a$ was taken from $x = 1$ instead of $x = 0$.

$P(x) = 90(3)^x$: Each value was divided by the next value instead of the previous value, giving $b = 3$.

$P(x) = 30(3)^x$: The value of $a$ was taken from $x = 1$, and the common factor was found by dividing each value by the next one.

$A(t) = 2{,}500(1.005)^t$

Use $A(t) = ab^t$. The initial value is $a = 2{,}500$. A 0.5% increase each year gives a growth factor of $b = 1 + 0.005 = 1.005$, so $A(t) = 2{,}500(1.005)^t$.

$A(t) = 2{,}500(0.50)^t$: The growth rate, 0.5%, was used as the growth factor without converting the percent.

$A(t) = 2{,}500 + 0.50t$: This linear function adds a constant amount; interest that grows by a percent is exponential.

$A(t) = 2{,}500(1.005)t$: The growth factor was multiplied by $t$ instead of raised to the power $t$.

$g(x) = 4\left(\frac{2}{5}\right)^x$

The $y$-intercept is $(0, 4)$, so $a = 4$. Dividing consecutive values gives $b = \frac{4}{10} = \frac{2}{5}$, so $g(x) = 4\left(\frac{2}{5}\right)^x$.

$g(x) = 4\left(\frac{1}{2}\right)^x$: This assumes the graph passes through $(1, 2)$, giving $b = \frac{1}{2}$.

$g(x) = \frac{2}{5}(4)^x$: The values of $a$ and $b$ were transposed.

$g(x) = \frac{1}{2}(4)^x$: This assumes the graph passes through $(1, 2)$ and transposes $a$ and $b$.

$f(x) = 4\left(\frac{1}{4}\right)^x$

Each value divided by the previous value is $\frac{16}{64} = \frac{1}{4}$, so $b = \frac{1}{4}$. The value when $x = 0$ is 4, so $a = 4$ and $f(x) = 4\left(\frac{1}{4}\right)^x$.

$f(x) = \frac{1}{4}(4)^x$: The values of $a$ and $b$ were reversed.

$f(x) = \left(\frac{1}{4}\right)^x$: The value of $a$ was taken from $x = 1$ instead of $x = 0$.

$f(x) = 4^x$: Each value was divided by the next value, giving $b = 4$, and $a$ was taken from $x = 1$.