c) \( y=-2 e^{3 x}+\sqrt{e} \)

To analyze the function \( y = -2 e^{3x} + \sqrt{e} \), we can explore its features such as its domain, range, intercepts, and asymptotic behavior. ### Step 1: Determine the Domain The domain of the function is the set of all possible values of \( x \) for which the function is defined. Since the exponential function \( e^{3x} \) is defined for all real numbers, the domain of \( y \) is: \[ \text{Domain: } (-\infty, \infty) \] ### Step 2: Determine the Range To find the range, we need to analyze the behavior of the function as \( x \) approaches positive and negative infinity. - As \( x \to -\infty \): \[ e^{3x} \to 0 \implies y \to -2(0) + \sqrt{e} = \sqrt{e} \] - As \( x \to \infty \): \[ e^{3x} \to \infty \implies y \to -2(\infty) + \sqrt{e} = -\infty \] Thus, the function decreases from \( \sqrt{e} \) to \( -\infty \). Therefore, the range is: \[ \text{Range: } (-\infty, \sqrt{e}] \] ### Step 3: Find the Intercepts #### y-intercept: To find the y-intercept, set \( x = 0 \): \[ y(0) = -2 e^{3(0)} + \sqrt{e} = -2(1) + \sqrt{e} = \sqrt{e} - 2 \] #### x-intercept: To find the x-intercept, set \( y = 0 \): \[ 0 = -2 e^{3x} + \sqrt{e} \] \[ 2 e^{3x} = \sqrt{e} \] \[ e^{3x} = \frac{\sqrt{e}}{2} \] Taking the natural logarithm of both sides: \[ 3x = \ln\left(\frac{\sqrt{e}}{2}\right) \] \[ x = \frac{1}{3} \ln\left(\frac{\sqrt{e}}{2}\right) \] ### Step 4: Asymptotic Behavior As \( x \to -\infty \), \( y \to \sqrt{e} \) (horizontal asymptote). As \( x \to \infty \), \( y \to -\infty \). ### Summary of Features - **Domain:** \( (-\infty, \infty) \) - **Range:** \( (-\infty, \sqrt{e}] \) - **y-intercept:** \( \sqrt{e} - 2 \) - **x-intercept:** \( \frac{1}{3} \ln\left(\frac{\sqrt{e}}{2}\right) \) - **Horizontal asymptote:** \( y = \sqrt{e} \) If you need further analysis or specific calculations, please let me know!