Answer:
The standard equation of parabola:
\(y =a(x-h)^2+k\) ....[1]
where, vertex = (h, k) and a is any constant value.
As per the statement:
In the given graph of parabola:
Vertex of parabola = (-2, -4)
⇒h = -2 and k= -4
Substitute in [1] we have;
\(y =a(x+2)^2-4\) ....[2]
Since, the graph of parabola cuts at y-axis at (0, -8)
Substitute this point in [2] to solve for a, we have;'
\(-8=a(0+2)^2-4\)
⇒\(-8=a(2)^2-4\)
Add 4 to both sides we have;
\(-4=a(2)^2\)
⇒\(-4=4a\)
Divide both sides by 4 we have;
a = -1
then, the equation become:
\(y =-1(x+2)^2-4\)
Therefore, the equation of the graph is, \(y =-(x+2)^2-4\)
