The given function is $f(x) = 2x^2 - 4x + 1$.

Key Features of the Quadratic Function:

1. Standard Form: The function is already in the standard quadratic form: $f(x) = ax^2 + bx + c$ where $a = 2$, $b = -4$, and $c = 1$.

2. Vertex: The vertex of a quadratic function $f(x) = ax^2 + bx + c$ can be found using the formula: $x_{\text{vertex}} = \frac{-b}{2a}$ Substituting $a = 2$ and $b = -4$: $x_{\text{vertex}} = \frac{-(-4)}{2(2)} = \frac{4}{4} = 1$ To find the y-coordinate of the vertex, substitute $x = 1$ into the function: $f(1) = 2(1)^2 - 4(1) + 1 = 2 - 4 + 1 = -1$ So, the vertex is $(1, -1)$.

3. Axis of Symmetry: The axis of symmetry is the vertical line that passes through the vertex. Thus, it is: $x = 1$

4. Direction of Opening: Since the coefficient of $x^2$ is positive (i.e., $a = 2$), the parabola opens upward.

5. Y-intercept: The y-intercept is the value of $f(x)$ when $x = 0$: $f(0) = 2(0)^2 - 4(0) + 1 = 1$ So, the y-intercept is $(0, 1)$.

Summary:

• Vertex: $(1, -1)$
• Axis of Symmetry: $x = 1$
• Opens upward (parabola)
• Y-intercept: $(0, 1)$

Would you like further details or have any questions about this function?

Here are 5 questions to explore further:

1. How do we find the x-intercepts of this quadratic function?
2. What is the domain and range of $f(x) = 2x^2 - 4x + 1$?
3. Can you graph the function using the key points found?
4. How would the function change if $a$ were negative?
5. How do we complete the square to rewrite this function in vertex form?

Tip: The vertex form of a quadratic function can make it easier to identify the vertex and understand transformations.