Answer:
The answer in the procedure
Step-by-step explanation:
The question does not present the graph, however it can be answered to help the student solve similar problems.
we know that
The equation of a vertical parabola into vertex form is equal to
[tex]f(x)=a(x-h)^{2}+k[/tex]
where
a is a coefficient
(h,k) is the vertex
If the coefficient a is positive then the parabola open up and the vertex is a minimum
If the coefficient a is negative then the parabola open down and the vertex is a maximum
case A) we have
[tex]f(x)=3(x+4)^{2}-6[/tex]
The vertex is the point (-4,-6)
a=3
therefore
The parabola open up, the vertex is a minimum
case B) we have
[tex]f(x)=3(x+4)^{2}-38[/tex]
The vertex is the point (-4,-38)
a=3
therefore
The parabola open up, the vertex is a minimum
case C) we have
[tex]f(x)=3(x-4)^{2}-6[/tex]
The vertex is the point (4,-6)
a=3
therefore
The parabola open up, the vertex is a minimum
case D) we have
[tex]f(x)=3(x-4)^{2}-38[/tex]
The vertex is the point (4,-38)
a=3
therefore
The parabola open up, the vertex is a minimum
