1. **State the problem:** We are given the function $$V(t) = 10 + \frac{20t + 5}{\sqrt{4t^2 + 9}}$$ which represents voltage as a function of time $t$. We want to understand and analyze this function. 2. **Formula and rules:** The function consists of a constant term 10 plus a rational expression where the numerator is linear in $t$ and the denominator is the square root of a quadratic expression in $t$. Important rules: - The denominator $\sqrt{4t^2 + 9}$ is always positive since $4t^2 + 9 > 0$ for all real $t$. - The function is defined for all real $t$. 3. **Simplify and analyze:** - The denominator can be written as $\sqrt{4t^2 + 9} = \sqrt{(2t)^2 + 3^2}$. - The numerator is $20t + 5$. 4. **Behavior at large $|t|$:** - For large $t$, $\sqrt{4t^2 + 9} \approx 2|t|$. - So, $$\frac{20t + 5}{\sqrt{4t^2 + 9}} \approx \frac{20t + 5}{2|t|}.$$ - When $t \to +\infty$, this approximates to $$\frac{20t}{2t} = 10,$$ so $$V(t) \approx 10 + 10 = 20.$$ - When $t \to -\infty$, $$\frac{20t}{2|t|} = \frac{20t}{-2t} = -10,$$ so $$V(t) \approx 10 - 10 = 0.$$ 5. **Evaluate at $t=0$:** - $$V(0) = 10 + \frac{20\cdot0 + 5}{\sqrt{4\cdot0^2 + 9}} = 10 + \frac{5}{3} = \frac{35}{3} \approx 11.67.$$ 6. **Summary:** - The function $V(t)$ is continuous and defined for all real $t$. - It approaches 20 as $t \to +\infty$ and 0 as $t \to -\infty$. - At $t=0$, $V(0) \approx 11.67$. **Final answer:** The voltage function is $$V(t) = 10 + \frac{20t + 5}{\sqrt{4t^2 + 9}}$$ with behavior described above.