Some point of interest might be the zeros and the vertex. By inspection alone the zeros appear to be -2.4 and .8, and the vertex is ( -1 , -5 ).

Next overlay a new graph replacing each x by ( x - 4 ).

The parabola moved to the right 4 units. The vertex doesn't appear to move up or down but slide along the line y= -5.

The next task is to get the parabola into the second quadrant. Since replacing x with ( x - 4 ) moved the parabola to the left, if we want the parabola in the second quadrant we will need to get the vertex to move up and to the left. This can be accomplished by replacing x by ( x + 4 ) and adding more than 5 to the end of the equation.

This graph is contained in the second quadrant for x< 0 . When the values of x are greater than zero part of the graph will lie in the first quadrant.

To have the parabola open down with the same vertex the exact coordinate of the vertex is a must. Using the completing the square technique the vertex is ( -3/4, -41/8 ). To see the work click here.In verex form the original eqauation looks like:

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Now if we change 2 to -2   the parabola will open down and have the same vertex as the orignial.

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The expanded version of the red graph is

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