Question

Find the coordinates of the vertex of the following parabola algebraically. Write your answer as an (x, y) point. y = x2 + 12x + 40

Answer 1

The given equation of parabola:

`y=x^2+12x+40`

The general equation of Parabola is:

`\begin{gathered} y=ax^2+bx+c \\ \text{Here the x coordinate of the vertex of given parabola is: }x=-(b)/(2a) \end{gathered}`

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,

`\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}`

The value of y-coordinate is 4

So, Vertex of parabola (x, y) : (-6, 4)

Answer: coordinates of vertex: (-6,4)