Question

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 $x=0$, $f(0)=84$ into $y=ab^x$:
$$84 = ab^0$$
Since $b^0=1$, we get $a=84$.

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

Substitute $a=84$, $x=-5$, $f(-5)=12$ into $y=ab^x$:
$$12 = 84b^{-5}$$
Rearrange to solve for $b^{-5}$:
$$b^{-5} = \frac{12}{84} = \frac{1}{7}$$
Rewrite as $b^5=7$, so $b = 7^{\frac{1}{5}}$

Step3: Write full function formula

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

Step4: Calculate $f(-3.5)$

Substitute $x=-3.5$:
$$f(-3.5) = 84 \times 7^{\frac{-3.5}{5}} = 84 \times 7^{-0.7}$$
Calculate $7^{-0.7} = \frac{1}{7^{0.7}} \approx \frac{1}{3.959} \approx 0.2526$
Then $84 \times 0.2526 \approx 21.2$

Answer:

21.2