To complete the square for the quadratic equation [tex]\( y = x^2 - 6x + 15 \)[/tex] and rewrite it in vertex form, follow these steps:
1. Quadratic Equation: Start with the given equation:
[tex]\[
y = x^2 - 6x + 15
\][/tex]
2. Identifying the Coefficient: The coefficient of the [tex]\( x \)[/tex] term is [tex]\(-6\)[/tex].
3. Completing the Square:
- Take half of the coefficient of [tex]\( x \)[/tex], which is [tex]\(-6\)[/tex], divide it by 2:
[tex]\[
\frac{-6}{2} = -3
\][/tex]
- Square this result:
[tex]\[
(-3)^2 = 9
\][/tex]
4. Rewriting the Equation: Add and subtract this square ([tex]\( 9 \)[/tex]) inside the equation:
[tex]\[
y = x^2 - 6x + 9 - 9 + 15
\][/tex]
Combine the terms inside the square and the constants:
[tex]\[
y = (x - 3)^2 - 9 + 15
\][/tex]
5. Simplify: Combine the constants [tex]\(-9\)[/tex] and [tex]\(15\)[/tex]:
[tex]\[
y = (x - 3)^2 + 6
\][/tex]
Now, the quadratic equation is in the vertex form [tex]\( y = a(x - h)^2 + k \)[/tex], where [tex]\( (h, k) \)[/tex] is the vertex.
Vertex form:
[tex]\[
y = (x - 3)^2 + 6
\][/tex]
From this, we see the vertex [tex]\((h, k)\)[/tex] is:
[tex]\[
(3, 6)
\][/tex]
6. Determining Maximum or Minimum:
- Since the coefficient of [tex]\( (x - 3)^2 \)[/tex] is positive (1), the parabola opens upwards.
- Therefore, the vertex represents the minimum point of the parabola.
Thus, the vertex of the quadratic function is at [tex]\( (3, 6) \)[/tex], and it is a minimum.
The correct answer is:
D. Minimum at [tex]\( (3, 6) \)[/tex]
