Answer with explanation:
We know that the general equation of a parabola in vertex form is given by:
[tex]y=a(x-h)^2+k[/tex]
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)
[tex]f(x)=-2(x+3)^2-1[/tex]
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)
[tex]f(x)=-2(x+3)^2+1[/tex]
Again the leading coefficient is negative.
Hence, graph is open downward.
The vertex of the function is at (-3,1)
c)
[tex]f(x)=2(x+3)^2+1[/tex]
The leading coefficient is positive.
Hence, graph is open upward.
The vertex of the function is at (-3,1)
d)
[tex]f(x)=2(x-3)^2+1[/tex]
The leading coefficient is positive.
Hence, graph is open upward.
The vertex of the function is at (3,1)
