All of the given equations have the following general form:
[tex]f(x)=a(x+h)^2+k[/tex]To analyze the equations and find their corresponding graph, we need to remember the following:
• If a>0, the parabola will open upwards, and if a<0, the parabola will open downwards.
,• The point (-h,k) is the vertex of the parabola.
Let's start with the first function:
[tex]f(x)=-2(x+3)^2-1[/tex]And identify a, h and k:
[tex]\begin{gathered} a=-2 \\ h=3 \\ k=-1 \end{gathered}[/tex]since a<0, the parabola will open downwards (this discards the first and third graphs since they open upwards).
And the vertex of the parabola is at:
[tex](-h,k)\longrightarrow(-3,-1)[/tex]And as we can see, the fourth graph is the one that opens downwards and has its vertex at (-3,-1):
We continue with the second function:
[tex]f(x)=-2(x+3)^2+1[/tex]It is very similar to the first option. Since a<0, the parabola opens downwards. And the vertex is at:
[tex]\begin{gathered} h=3 \\ k=1 \\ Vertex\colon(-h,k)\longrightarrow(-3,1) \end{gathered}[/tex]The second graph is the one that meets these characteristics:
Now, we analyze the third function:
[tex]f(x)=2(x+3)^2+1[/tex]And identify a, h and k:
[tex]\begin{gathered} a=2 \\ h=3 \\ k=1 \end{gathered}[/tex]Since a>0, the parabola will open upwards. Now we find the vertex point to decide if this is represented by the first or the third graph:
[tex](-h,k)\longrightarrow(-3,1)[/tex]These characteristics coincide with the first graph:
Finally, we analyze the fourth function:
[tex]f(x)=2(x-3)^2+1[/tex]Identify a, h an k:
[tex]undefined[/tex]Again a>0, thus the parabola will open upwards.
