Is y=8^x-6 an increasing or decreasing function? square 4 Is y=2((1)/(2))^-x+2 an increasing or decreasing function? square

Let's analyze each function separately to determine whether they are increasing or decreasing.<br /><br />### Function 1: \( y = 8^x - 6 \)<br /><br />To determine if this function is increasing or decreasing, we need to look at the base of the exponential term, which is 8 in this case.<br /><br />- If the base of the exponential term is greater than 1, the function is increasing.<br />- If the base of the exponential term is between 0 and 1, the function is decreasing.<br /><br />Since \( 8 > 1 \), the function \( y = 8^x - 6 \) is an increasing function.<br /><br />### Function 2: \( y = 2 \left( \frac{1}{2} \right)^{-x} + 2 \)<br /><br />First, let's rewrite the function for clarity:<br />\[ y = 2 \left( \frac{1}{2} \right)^{-x} + 2 \]<br />\[ y = 2 \cdot 2^x + 2 \]<br />\[ y = 2^{x+1} + 2 \]<br /><br />Now, let's analyze the base of the exponential term, which is \( 2 \).<br /><br />- If the base of the exponential term is greater than 1, the function is increasing.<br />- If the base of the exponential term is between 0 and 1, the function is decreasing.<br /><br />Since \( 2 > 1 \), the function \( y = 2^{x+1} + 2 \) is an increasing function.<br /><br />### Summary<br /><br />- The function \( y = 8^x - 6 \) is an increasing function.<br />- The function \( y = 2 \left( \frac{1}{2} \right)^{-x} + 2 \) is an increasing function.<br /><br />So, the answers are:<br /><br />1. \( y = 8^x - 6 \) is an increasing function.<br />2. \( y = 2 \left( \frac{1}{2} \right)^{-x} + 2 \) is an increasing function.