1. **State the problem:** We are given the quadratic function $f(x) = x^2 + 4x + 5$ and want to analyze it. 2. **Formula and rules:** The general form of a quadratic function is $f(x) = ax^2 + bx + c$ where $a=1$, $b=4$, and $c=5$ here. 3. **Find the vertex:** The vertex form is useful to find the minimum or maximum point. The $x$-coordinate of the vertex is given by $$x = -\frac{b}{2a} = -\frac{4}{2 \times 1} = -2.$$ 4. **Calculate the $y$-coordinate of the vertex:** Substitute $x=-2$ into $f(x)$: $$f(-2) = (-2)^2 + 4(-2) + 5 = 4 - 8 + 5 = 1.$$ 5. **Rewrite in vertex form:** $$f(x) = (x + 2)^2 + 1.$$ 6. **Interpretation:** Since $a=1 > 0$, the parabola opens upwards and the vertex at $(-2,1)$ is the minimum point. 7. **Find the $x$-intercepts:** Solve $x^2 + 4x + 5 = 0$ using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{16 - 20}}{2} = \frac{-4 \pm \sqrt{-4}}{2}.$$ Since the discriminant is negative, there are no real $x$-intercepts. 8. **Find the $y$-intercept:** Set $x=0$: $$f(0) = 0 + 0 + 5 = 5.$$ **Final answer:** The vertex is at $(-2,1)$, the parabola opens upward, there are no real $x$-intercepts, and the $y$-intercept is at $(0,5)$.