The given equation of parabola:
[tex]y=x^2+12x+40[/tex]
The general equation of Parabola is:
[tex]\begin{gathered} y=ax^2+bx+c \\ \text{Here the x coordinate of the vertex of given parabola is: }x=-\frac{b}{2a} \end{gathered}[/tex]
On comparing the general equation with the given equation we get:
a = 1, b = 12 and c = 40
So, the x -coordinate of vertex is : x = -12/2(1)
x = -6
Substitute the value of x = -6 in the expression of Parabola to find the value of y -coordinate
So,
[tex]\begin{gathered} y=x^2+12x+40 \\ \text{Substitute x = -6} \\ y=(-6)^2+12(-6)+40 \\ y=36-72+40 \\ y=4 \end{gathered}[/tex]
The value of y-coordinate is 4
So, Vertex of parabola (x, y) : (-6, 4)
Answer: coordinates of vertex: (-6,4)
