To rewrite the quadratic function $f(x) = x^2 - 8x + 15$ in vertex form, we need to complete the square.

The general form of a quadratic function in vertex form is:

$f(x) = a(x - h)^2 + k$

where $(h, k)$ is the vertex of the parabola.

Steps to convert the given quadratic to vertex form:

1. Start with the quadratic:
   $f(x) = x^2 - 8x + 15$

2. Group the $x$-terms:
   We need to complete the square for the terms $x^2 - 8x$.
   $f(x) = (x^2 - 8x) + 15$

3. Complete the square:
   To complete the square, take half of the coefficient of $x$, square it, and add it and subtract it inside the parentheses.

   • The coefficient of $x$ is $-8$. Half of $-8$ is $-4$, and squaring it gives $16$.
     $f(x) = (x^2 - 8x + 16) - 16 + 15$

4. Rewrite as a perfect square:
   The expression $x^2 - 8x + 16$ is a perfect square trinomial, so we can factor it as: $f(x) = (x - 4)^2 - 1$

Thus, the vertex form of the function is:

$f(x) = (x - 4)^2 - 1$

Conclusion:

• The vertex form of $f(x) = x^2 - 8x + 15$ is $f(x) = (x - 4)^2 - 1$.
• The vertex of the parabola is $(4, -1)$.

Would you like to see the graph of this function or have further details?

Related questions:

1. How can you identify the vertex from the equation in standard form?
2. How do the values of $a$, $h$, and $k$ in the vertex form affect the graph of the quadratic function?
3. What is the axis of symmetry for the quadratic function $f(x) = (x - 4)^2 - 1$?
4. How can you find the vertex and rewrite the function if the quadratic has no constant term?
5. How do you convert from vertex form back to standard form?

Tip: Always remember that completing the square involves adding and subtracting the same number, which ensures the expression remains equivalent to the original.