To find the vertex of the quadratic function $f(x) = 4x^2 - 16x + 10$, we can use the vertex formula.

For a quadratic function in the form $f(x) = ax^2 + bx + c$, the x-coordinate of the vertex is given by:

$x = -\frac{b}{2a}$

Here, the coefficients are:

• $a = 4$
• $b = -16$
• $c = 10$

Let's calculate the x-coordinate of the vertex:

$x = -\frac{-16}{2 \times 4} = \frac{16}{8} = 2$

Now that we have the x-coordinate, we can substitute it back into the function to find the y-coordinate:

$f(2) = 4(2)^2 - 16(2) + 10 = 4(4) - 32 + 10 = 16 - 32 + 10 = -6$

So, the vertex of the function $f(x) = 4x^2 - 16x + 10$ is at the point $(2, -6)$.

Do you want more details or have any questions?

Here are 5 related questions that could help you expand your understanding:

1. How can you determine if the vertex is a maximum or minimum point?
2. How would you complete the square for the quadratic function to find the vertex?
3. What does the graph of $f(x) = 4x^2 - 16x + 10$ look like?
4. Can you find the roots of the quadratic function using the vertex form?
5. How does changing the coefficient $a$ affect the position of the vertex?

Tip: The vertex form of a quadratic equation is another useful method for finding the vertex directly. The vertex form is $f(x) = a(x-h)^2 + k$, where $(h, k)$ is the vertex.