Question

if $f(x)$ is an exponential function of the form of $y = ab^{x}$ where $f(-5) = 12$ and $f(0) = 84$, then find the value of $f(-3.5)$, to the nearest tenth.

Explanation:

Step1: Find $a$ using $f(0)=84$

Substitute $x=0$ into $y=ab^x$:
$f(0)=ab^0 = a(1) = a = 84$

Step2: Find $b$ using $f(-5)=12$

Substitute $a=84$, $x=-5$, $f(-5)=12$:
$12 = 84b^{-5}$
Rearrange to solve for $b^5$:
$b^{-5} = \frac{12}{84} = \frac{1}{7}$
$b^5 = 7$
Solve for $b$:
$b = 7^{\frac{1}{5}}$

Step3: Write full function

$f(x) = 84 \cdot (7^{\frac{1}{5}})^x = 84 \cdot 7^{\frac{x}{5}}$

Step4: Calculate $f(-3.5)$

Substitute $x=-3.5$:
$f(-3.5) = 84 \cdot 7^{\frac{-3.5}{5}} = 84 \cdot 7^{-0.7}$
Calculate $7^{-0.7} = \frac{1}{7^{0.7}} \approx \frac{1}{3.899}$
$f(-3.5) \approx 84 \cdot 0.2565 \approx 21.5$

Answer:

$21.5$