1. **State the problem:** We need to analyze the function $$y=\sqrt{5x^2+1}$$. 2. **Formula and rules:** The function is a square root of a quadratic expression. The square root function is defined only for values where the expression inside is non-negative. 3. **Domain:** Since $$5x^2+1 \geq 0$$ for all real $$x$$ (because $$5x^2 \geq 0$$ and adding 1 keeps it positive), the domain is all real numbers. 4. **Simplify and analyze:** The function can be written as $$y=(5x^2+1)^{\frac{1}{2}}$$. 5. **Find intercepts:** - **y-intercept:** Set $$x=0$$, then $$y=\sqrt{5\cdot0^2+1}=\sqrt{1}=1$$. - **x-intercept:** Set $$y=0$$, then $$\sqrt{5x^2+1}=0$$ implies $$5x^2+1=0$$ which has no real solution. 6. **Find extrema:** - Take derivative: $$y' = \frac{1}{2}(5x^2+1)^{-\frac{1}{2}} \cdot 10x = \frac{5x}{\sqrt{5x^2+1}}$$. - Set $$y'=0$$ to find critical points: $$5x=0 \Rightarrow x=0$$. - Second derivative test or analyze sign change: - For $$x<0$$, $$y'<0$$ (decreasing), for $$x>0$$, $$y'>0$$ (increasing), so $$x=0$$ is a minimum. - Minimum value at $$x=0$$ is $$y=1$$. **Final answer:** The function $$y=\sqrt{5x^2+1}$$ has domain all real numbers, a minimum at $$(0,1)$$, no x-intercepts, and y-intercept at $$(0,1)$$.