To find the vertex of the quadratic function [tex]\( f(x) = x^2 + 6x + 25 \)[/tex], we follow these steps:
1. Identify the coefficients from the given quadratic function [tex]\( ax^2 + bx + c \)[/tex]:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = 6 \ )
- \( c = 25 \)[/tex]
2. Find the x-coordinate of the vertex:
- The x-coordinate of the vertex for a quadratic equation in standard form is given by the formula:
[tex]\[
x = -\frac{b}{2a}
\][/tex]
- Substitute the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[
x = -\frac{6}{2 \times 1} = -\frac{6}{2} = -3
\][/tex]
3. Find the y-coordinate of the vertex:
- To obtain the y-coordinate, substitute the x-coordinate back into the original quadratic equation:
[tex]\[
y = f(-3) = (-3)^2 + 6(-3) + 25
\][/tex]
- Calculate each term:
[tex]\[
(-3)^2 = 9
\][/tex]
[tex]\[
6(-3) = -18
\][/tex]
[tex]\[
y = 9 - 18 + 25
\][/tex]
[tex]\[
y = -9 + 25
\][/tex]
[tex]\[
y = 16
\][/tex]
Thus, the vertex of the quadratic function [tex]\( f(x) = x^2 + 6x + 25 \)[/tex] is at [tex]\( (-3, 16) \)[/tex].
