1. The problem is to analyze and understand the quadratic function $f(x) = 2x^2 + 4x + 1$ and describe its graph. 2. The general form of a quadratic function is $f(x) = ax^2 + bx + c$, where $a$, $b$, and $c$ are constants. Here, $a=2$, $b=4$, and $c=1$. 3. Since $a=2 > 0$, the parabola opens upwards. 4. To find the vertex, use the formula for the $x$-coordinate of the vertex: $$x = -\frac{b}{2a} = -\frac{4}{2 \times 2} = -\frac{4}{4} = -1.$$ 5. Substitute $x=-1$ into $f(x)$ to find the $y$-coordinate of the vertex: $$f(-1) = 2(-1)^2 + 4(-1) + 1 = 2(1) - 4 + 1 = 2 - 4 + 1 = -1.$$ 6. So, the vertex is at $(-1, -1)$. 7. To find the $y$-intercept, evaluate $f(0)$: $$f(0) = 2(0)^2 + 4(0) + 1 = 1.$$ 8. To find the $x$-intercepts, solve $2x^2 + 4x + 1 = 0$ using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{16 - 8}}{4} = \frac{-4 \pm \sqrt{8}}{4} = \frac{-4 \pm 2\sqrt{2}}{4} = \frac{\cancel{-4} \pm 2\sqrt{2}}{\cancel{4}} = -1 \pm \frac{\sqrt{2}}{2}.$$ 9. The $x$-intercepts are $x = -1 + \frac{\sqrt{2}}{2}$ and $x = -1 - \frac{\sqrt{2}}{2}$. 10. The parabola is symmetric about the vertical line $x = -1$. Final answer: The quadratic function $f(x) = 2x^2 + 4x + 1$ is an upward-opening parabola with vertex at $(-1, -1)$, $y$-intercept at $(0,1)$, and $x$-intercepts at $\left(-1 + \frac{\sqrt{2}}{2}, 0\right)$ and $\left(-1 - \frac{\sqrt{2}}{2}, 0\right)$.