I need help with this please

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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;

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