How does \( h(x)=5 \cdot 0.75^{x} \) change over the interval from \( x=-8 \) to \( x=-7 \) ? \( h(x) \) decreases by \( 25 \% \) \( h(x) \) increases by \( 175 \% \) \( h(x) \) decreases by \( 75 \% \) \( h(x) \) decreases by 0.75

To determine how the function \( h(x) = 5 \cdot 0.75^{x} \) changes as \( x \) increases from \(-8\) to \(-7\), let's analyze the behavior step by step. 1. **Understanding the Function Behavior:** The function \( h(x) = 5 \cdot 0.75^{x} \) is an exponential function where the base \( 0.75 \) is less than 1. Generally, for exponential functions of the form \( y = A \cdot b^x \) where \( 0 < b < 1 \), the function decreases as \( x \) increases. 2. **Calculating the Change from \( x = -8 \) to \( x = -7 \):** - **At \( x = -8 \):** \[ h(-8) = 5 \cdot 0.75^{-8} = 5 \cdot \left(\frac{1}{0.75}\right)^8 = 5 \cdot \left(\frac{4}{3}\right)^8 \] (The exact value isn't necessary for understanding the percentage change.) - **At \( x = -7 \):** \[ h(-7) = 5 \cdot 0.75^{-7} = 5 \cdot \left(\frac{1}{0.75}\right)^7 = 5 \cdot \left(\frac{4}{3}\right)^7 \] - **Comparing \( h(-7) \) to \( h(-8) \):** \[ h(-7) = 0.75 \cdot h(-8) \] This means \( h(-7) \) is \( 75\% \) of \( h(-8) \). 3. **Interpreting the Percentage Change:** Since \( h(-7) \) is \( 75\% \) of \( h(-8) \), the function has **decreased by \( 25\% \)** over the interval from \( x = -8 \) to \( x = -7 \). **Answer:** \( h(x) \) decreases by \( 25 \% \).