we know that
The equation of a vertical parabola in vertex form is equal to

where
(h,k) is the vertex
if
----> the parabola open upward (vertex is a minimum)
if
----> the parabola open downward (vertex is a maximum)
case A)

In this problem we have
the vertex is the point


therefore
the parabola open downward (vertex is a maximum)
The graph with its corresponding equation in the attached figure
case B)

convert to vertex form
Factor the leading coefficient

Complete the square. Remember to balance the equation by adding the same constants to each side.

Rewrite as perfect squares


In this problem we have
the vertex is the point


therefore
the parabola open upward (vertex is a minimum)
The graph with its corresponding equation in the attached figure
case C)

convert to vertex form
Group terms that contain the same variable, and move the constant to the opposite side of the equation

Complete the square. Remember to balance the equation by adding the same constants to each side.


Rewrite as perfect squares


In this problem we have
the vertex is the point


the y-intercept is

remember that the y-intercept is the value of y when the value of x is equal to zero
therefore
the parabola open downward (vertex is a maximum)
The graph with its corresponding equation is not in the attached figure
case D)

convert to vertex form
Group terms that contain the same variable, and move the constant to the opposite side of the equation

Factor the leading coefficient
Complete the square. Remember to balance the equation by adding the same constants to each side.

Rewrite as perfect squares

In this problem we have
the vertex is the point


therefore
the parabola open upward (vertex is a minimum)
The graph with its corresponding equation in the attached figure
case E)

convert to vertex form
Group terms that contain the same variable, and move the constant to the opposite side of the equation

Factor the leading coefficient

Complete the square. Remember to balance the equation by adding the same constants to each side.
Rewrite as perfect squares
In this problem we have
the vertex is the point


the y-intercept is

remember that the y-intercept is the value of y when the value of x is equal to zero
therefore
the parabola open downward (vertex is a maximum)
The graph with its corresponding equation in the attached figure
case F)

In this problem we have
the vertex is the point

therefore
the parabola open upward (vertex is a minimum)
The graph with its corresponding equation is not in the attached figure
therefore
the answer in the attached figure