Sure! Let’s start analyzing the quadratic function $f(x) = 4x^2 - 4x + 2$. Here’s how we can approach answering various questions related to this function:

1. Determine the vertex of the quadratic function.

The vertex form of a quadratic function is given by: $f(x) = a(x - h)^2 + k$ where $(h, k)$ is the vertex.

We can find the vertex by using the formula for the x-coordinate of the vertex: $x = \frac{-b}{2a}$ where $a = 4$ and $b = -4$ from the given function.

$x = \frac{-(-4)}{2(4)} = \frac{4}{8} = \frac{1}{2}$

Now, substitute $x = \frac{1}{2}$ back into the function to find the y-coordinate: $f\left(\frac{1}{2}\right) = 4\left(\frac{1}{2}\right)^2 - 4\left(\frac{1}{2}\right) + 2$ $f\left(\frac{1}{2}\right) = 4\left(\frac{1}{4}\right) - 2 + 2 = 1$

Thus, the vertex is $\left( \frac{1}{2}, 1 \right)$.

2. Find the axis of symmetry.

The axis of symmetry is the vertical line that passes through the vertex. Since the x-coordinate of the vertex is $\frac{1}{2}$, the axis of symmetry is: $x = \frac{1}{2}$

3. Determine whether the function has a minimum or maximum value.

Since the coefficient of $x^2$ (i.e., $a = 4$) is positive, the parabola opens upwards. This means the function has a minimum value at the vertex.

4. Find the minimum or maximum value.

The minimum value of the function is the y-coordinate of the vertex, which we calculated as $1$. Therefore, the minimum value of the function is: $f(x)_{\text{min}} = 1$

5. Find the y-intercept.

To find the y-intercept, set $x = 0$ and solve for $f(0)$: $f(0) = 4(0)^2 - 4(0) + 2 = 2$ Thus, the y-intercept is $(0, 2)$.

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Let me know if you'd like any more details or clarifications! Here are five related questions:

1. How do we convert the quadratic function into vertex form?
2. What are the x-intercepts of the function $f(x) = 4x^2 - 4x + 2$?
3. Can you graph the function based on the vertex, axis of symmetry, and intercepts?
4. How does the discriminant of a quadratic equation relate to the nature of its roots?
5. How do changes to the coefficient $a$ in $ax^2 + bx + c$ affect the shape of the parabola?

Tip: The vertex form of a quadratic function can be useful for quickly identifying the vertex and graphing the function.