<h2>
Answer with explanation:</h2>
We know that the general equation of a parabola in vertex form is given by:
![y=a(x-h)^2+k](https://tex.z-dn.net/?f=y%3Da%28x-h%29%5E2%2Bk)
where the vertex of the parabola is at (h,k)
and if a>0 then the parabola is open upward and if a<0 then the parabola is open downward.
a)
![f(x)=-2(x+3)^2-1](https://tex.z-dn.net/?f=f%28x%29%3D-2%28x%2B3%29%5E2-1)
Since, the leading coefficient is negative.
Hence, the graph of the function is a parabola which is downward open.
The vertex of the function is at (-3,-1)
b)
![f(x)=-2(x+3)^2+1](https://tex.z-dn.net/?f=f%28x%29%3D-2%28x%2B3%29%5E2%2B1)
Again the leading coefficient is negative.
Hence, graph is open downward.
The vertex of the function is at (-3,1)
c)
![f(x)=2(x+3)^2+1](https://tex.z-dn.net/?f=f%28x%29%3D2%28x%2B3%29%5E2%2B1)
The leading coefficient is positive.
Hence, graph is open upward.
The vertex of the function is at (-3,1)
d)
![f(x)=2(x-3)^2+1](https://tex.z-dn.net/?f=f%28x%29%3D2%28x-3%29%5E2%2B1)
The leading coefficient is positive.
Hence, graph is open upward.
The vertex of the function is at (3,1)