EXPLANATION
Given the algebraic expression f(x)=x^2 -12x + 44
In order to identify the vertex we need to apply the following equation:
[tex]\text{vertex}=x_v=-\frac{b}{2a}[/tex]
In this case, a=1, b=-12 and c=44
Replacing terms:
[tex]x_v=-\frac{-12}{2\cdot1}[/tex]
Simplifying:
[tex]x_v=\frac{12}{2}=6[/tex]
The vertex is at point x=6
Plug in x=6 to find the y_v value:
[tex]y_v=6^2-12\cdot6+44=36-72+44=8[/tex]
Therefore, the vertex of the parabola is:
Vertex= (6,8)
Minimum is (6,8) because a=1>0, and then the vertex is a minimum.
Axis of simmetry:
Rewrite the equation in the standard form:
[tex]4p(y-k)=(x-h)^2\text{ with vertex at (h,k) and a focal length p}[/tex][tex]4\frac{1}{4}(y-8)=(x-6)^2[/tex]
Therefore the parabola properties are (h,k)=(6,8) , p=1/4
Parabola is of the form 4p(y-k)=(x-h)^2 and is symmetric around the y-axis
Axis of simmetry is a line parallel to the y-axis wich intersct the vertex.
x=6
