If [tex] f(x) [/tex] is an exponential function of the form [tex] y = ab^x [/tex], where [tex] f(-3.5) = 10 [/tex] and [tex] f(0.5) = 89 [/tex], find the value of [tex] f(0) [/tex] to the nearest hundredth.

To find the value of [tex]\( f(0) \)[/tex] for an exponential function [tex]\( f(x) = a \cdot b^x \)[/tex] given the points [tex]\( f(-3.5) = 10 \)[/tex] and [tex]\( f(0.5) = 89 \)[/tex], follow these steps:

1. Set up the equations using the points provided:

- For [tex]\( f(-3.5) = 10 \)[/tex]:
[tex]\[
10 = a \cdot b^{-3.5}
\][/tex]
- For [tex]\( f(0.5) = 89 \)[/tex]:
[tex]\[
89 = a \cdot b^{0.5}
\][/tex]

2. Solve for [tex]\( b \)[/tex] by dividing the two equations:

Divide the equation for [tex]\( f(0.5) \)[/tex] by the equation for [tex]\( f(-3.5) \)[/tex]:
[tex]\[
\frac{89}{10} = \frac{a \cdot b^{0.5}}{a \cdot b^{-3.5}}
\][/tex]
Simplify the expression:
[tex]\[
\frac{89}{10} = b^{0.5 + 3.5} = b^4
\][/tex]

3. Calculate [tex]\( b \)[/tex]:

Find [tex]\( b \)[/tex] by taking the fourth root of both sides:
[tex]\[
b = \left(\frac{89}{10}\right)^{\frac{1}{4}} \approx 1.727
\][/tex]

4. Solve for [tex]\( a \)[/tex]:

Substitute [tex]\( b \)[/tex] back into one of the original equations to find [tex]\( a \)[/tex]. Let's use [tex]\( f(-3.5) = 10 \)[/tex]:
[tex]\[
10 = a \cdot b^{-3.5}
\][/tex]
Solve for [tex]\( a \)[/tex]:
[tex]\[
a = 10 \cdot b^{3.5}
\][/tex]
Substitute the value of [tex]\( b \)[/tex]:
[tex]\[
a \approx 10 \cdot (1.727)^{3.5} \approx 67.72
\][/tex]

5. Find [tex]\( f(0) \)[/tex]:

Since [tex]\( f(x) = a \cdot b^x \)[/tex], then:
[tex]\[
f(0) = a \cdot b^0 = a
\][/tex]
Therefore, the value of [tex]\( f(0) \)[/tex] is:
[tex]\[
f(0) \approx 67.72
\][/tex]

Thus, the value of [tex]\( f(0) \)[/tex], rounded to the nearest hundredth, is approximately [tex]\( 67.72 \)[/tex].