To find the vertex of the quadratic function [tex]\( f(x) = x^2 + 12x \)[/tex], we can use the vertex formula. A quadratic function of the form [tex]\( f(x) = ax^2 + bx + c \)[/tex] has its vertex at the point [tex]\((h, k)\)[/tex], where:
- [tex]\( h = -\frac{b}{2a} \)[/tex]
- [tex]\( k = f(h) \)[/tex]
For the given function [tex]\( f(x) = x^2 + 12x \)[/tex]:
- The coefficient [tex]\( a \)[/tex] (the coefficient of [tex]\( x^2 \)[/tex]) is [tex]\( 1 \)[/tex].
- The coefficient [tex]\( b \)[/tex] (the coefficient of [tex]\( x \)[/tex]) is [tex]\( 12 \)[/tex].
First, we find the x-coordinate of the vertex [tex]\( h \)[/tex]:
[tex]\[ h = -\frac{b}{2a} = -\frac{12}{2 \cdot 1} = -\frac{12}{2} = -6 \][/tex]
Next, we find the y-coordinate of the vertex [tex]\( k \)[/tex] by substituting [tex]\( h \)[/tex] back into the function:
[tex]\[ k = f(h) = f(-6) = (-6)^2 + 12(-6) \][/tex]
[tex]\[ = 36 - 72 = -36 \][/tex]
Therefore, the vertex of the function [tex]\( f(x) = x^2 + 12x \)[/tex] is:
[tex]\[ (-6, -36) \][/tex]
So, the correct answer is:
[tex]\[ \boxed{(-6, -36)} \][/tex]
