Question

1. label each graph by writing exponential growth or exponential decay in the blank.

$f(x) = a(1 + r)^x$, where $r > 0$ _______________
$f(x) = a(1 - r)^x$, where $r > 0$ _______________

1. complete the steps for finding the value of a car after 5 years of depreciation.

initial value of a car: $15,000 \quad$ decay factor: 12% per year $\quad$ time: 5 years
$f(x) = a(1 - r)^x \quad \quad \quad \quad$ write the function to model exponential decay.
$f(x) = \underline{\quad\quad} (1 - \underline{\quad\quad})^{\underline{\quad\quad}} \quad$ substitute values for $a$, $r$, and $x$.
$f(x) = \underline{\quad\quad} \quad \quad \quad \quad \quad \quad$ simplify.
the value of a $15,000 car after 5 years would be around $\underline{\quad\quad}$.

1. hannah invested $4,000 in a savings account that earned 2% interest compounded quarterly. she determined that if she does not withdraw or deposit any more money, the value of the account at the end of 3 years will be $4,244.83. what error did hannah make in her calculations? what will the account balance be after 3 years? explain.

Explanation:

Step1: Identificar la función de decaimiento exponencial

La fórmula para el decaimiento exponencial es \( f(x) = a(1 - r)^x \), donde \( a \) es el valor inicial, \( r \) es la tasa de decaimiento (en decimal) y \( x \) es el tiempo en años. Aquí, \( a = 15000 \), \( r = 0.12 \) (porque \( 12\% = \frac{12}{100} = 0.12 \)) y \( x = 5 \).

Step2: Sustituir los valores en la fórmula

Sustituimos \( a = 15000 \), \( r = 0.12 \) y \( x = 5 \) en la fórmula: \( f(x) = 15000(1 - 0.12)^5 \).

Step3: Simplificar la expresión

Primero, calculamos \( 1 - 0.12 = 0.88 \). Luego, elevamos \( 0.88 \) a la potencia de 5: \( 0.88^5 \approx 0.527731937 \). Luego, multiplicamos por 15000: \( 15000 \times 0.527731937 \approx 7915.98 \).

Answer:

Para la función de decaimiento exponencial:

• \( f(x) = 15000(1 - 0.12)^5 \) (sustituyendo \( a = 15000 \), \( r = 0.12 \), \( x = 5 \))
• Al simplificar, \( f(x) \approx 7916 \) (redondeado a la nearest dollar)

Así, los espacios en blanco se completan como:

• \( f(x) = \boldsymbol{15000} (1 - \boldsymbol{0.12})^{5} \)
• Simplificando, \( f(x) \approx \boldsymbol{7916} \) (o \( 7915.98 \) si se mantiene la precisión decimal)