If \( f(x) \) is an exponential function of the form of \( y=a b^{x} \) where \( f(-1.5)=12 \) and \( f(2)=79 \), then find the value of \( f(-0.5) \), to the nearest hundredth.

To find the values of the constants \( a \) and \( b \) in the exponential function \( f(x) = ab^x \), we can use the given values: 1. We have \( f(-1.5) = 12 \): \[ ab^{-1.5} = 12 \quad \text{(1)} \] 2. We have \( f(2) = 79 \): \[ ab^{2} = 79 \quad \text{(2)} \] We can divide equation (2) by equation (1) to eliminate \( a \): \[ \frac{ab^2}{ab^{-1.5}} = \frac{79}{12} \] \[ b^{2 + 1.5} = \frac{79}{12} \] \[ b^{3.5} = \frac{79}{12} \] Now we calculate \( b \): \[ b = \left( \frac{79}{12} \right)^{\frac{1}{3.5}} = \left( \frac{79}{12} \right)^{\frac{2}{7}} \quad \text{(using } \frac{1}{3.5} = \frac{2}{7} \text{)} \] Now calculate \( \frac{79}{12} \): \[ \frac{79}{12} \approx 6.5833 \] Next, we compute \( b \): \[ b \approx (6.5833)^{\frac{2}{7}} \approx 2.235 \] Next, we can substitute \( b \) back into one of the original equations to find \( a \). Let's use equation (1): \[ ab^{-1.5} = 12 \] \[ a \cdot (2.235)^{-1.5} = 12 \] \[ a = 12 \cdot (2.235)^{1.5} \] Calculating \( (2.235)^{1.5} \): \[ (2.235)^{1.5} \approx 3.348 \] Thus: \[ a \approx 12 \cdot 3.348 \approx 40.176 \] Now we have both \( a \) and \( b \): \[ a \approx 40.176 \] \[ b \approx 2.235 \] Finally, we want to find \( f(-0.5) \): \[ f(-0.5) = ab^{-0.5} \approx 40.176 \cdot (2.235)^{-0.5} \] Calculating \( (2.235)^{-0.5} \): \[ (2.235)^{-0.5} \approx 0.669 \] Now we can calculate \( f(-0.5) \): \[ f(-0.5) \approx 40.176 \cdot 0.669 \approx 26.94 \] Thus, the value of \( f(-0.5) \) to the nearest hundredth is: \[ \boxed{26.94} \]