Answer:
D
Step-by-step explanation:
The equations are similar and so if we compute the vertex we would see they are the same..
Let's try computing the maximum or minimum point of y= -3(x+2)^2 - 4 and y= 3(x+2)^2 - 4 ;
For y= 3(x+2)^2 - 4 ;
to compute the maximum or minimum point we find dy/dx of the expression equating it to 0; the value of x is determined and we substitute to the original expression for y.
If y is -ve we know it's a minimum graph and if y is +be it's a maximum graph.
From the foregoing;
For y= 3(x+2)^2 - 4 ;
dy/dx = 2× 3 ( x+2) ×1 = 6(x+2) = 6x + 12= 0
6x= -12=>x= -12/6= -2;
We substitute x= -2 in y =3(x+2)^2 - 4; y = 3(-2+2)^2-4 = -4
Since y = -4 is a hence y= 3(x+2)^2 - 4 is a minimum graph.
For y= -3(x+2)^2 - 4 ;
dy/dx = 2× -3 ( x+2) ×1 = -6(x+2) = -6x - 12= 0
-6x= 12=>x= -12/6= -2;
Substituting in the y= -3(x+2)^2 - 4 ;
We have;
We substitute x= -2 in y =-3(x+2)^2 - 4; y = -3(-2+2)^2-4 = -4
y= -4 ;
Since both graphs look like they are minimum that is the vertex is y= -4;
Let's explore a further step
The derivative of the previous derivative;
For y= 3(x+2)^2 - 4;
We means d/dy × [dx/dy] = d/dy [ 6x + 12 ] = 6
Hence y= 3(x+2)^2 - 4 is a maximum graph;
Similarly for y= -3(x+2)^2 - 4 ;
d/dy × [dx/dy]
d/dy [-6x-12 ] = -6
Hence y= -3(x+2)^2 - 4 is a minimum graph;
Conclusion: y= 3(x+2)^2 - 4 is a maximum graph and y= -3(x+2)^2 - 4 is a minimum graph;
