1. **State the problem:** Find the intercepts and extrema of the function $y = x^2 + 1$. 2. **Formula and rules:** - To find the $y$-intercept, evaluate $y$ at $x=0$. - To find the $x$-intercepts, solve $y=0$ for $x$. - To find extrema, compute the derivative $y'$, set it to zero, and solve for $x$. 3. **Find intercepts:** - $y$-intercept: $y(0) = 0^2 + 1 = 1$ - $x$-intercepts: Solve $x^2 + 1 = 0$ which gives $x^2 = -1$; no real solutions, so no $x$-intercepts. 4. **Find extrema:** - Derivative: $y' = 2x$ - Set $y' = 0$: $2x = 0$ so $x=0$ - Evaluate $y$ at $x=0$: $y(0) = 1$ - Second derivative test: $y'' = 2 > 0$, so $x=0$ is a minimum. **Final answer:** The function has a minimum at $(0,1)$, no $x$-intercepts, and a $y$-intercept at $(0,1)$.