Consider the graph of .

We know that this is going to be a parabola because this is a quadratic equation (degree = 2). Let’s graph it and see what happens.

https://www.desmos.com/calculator/oljfrj8e2m

Now let’s graph the following substitute (x-4) for each x in the equation. This changes the equation to

https://www.desmos.com/calculator/oljfrj8e2m

This substitution creates the graph of an equation which is shifted 4 units to the right of the original parabola. This can be determined by looking at the points in the original and the points in the second parabola. For instance, the vertex of the original parabola is (-0.75, -5.125), and the vertex of the second parabola is (3.25, -5.125).

To move the vertex into the second quadrant, we can start by substituting (x+4) into the original equation. This will translate the parabola 4 units to the left, placing the vertex into the third quadrant.

Now, since the vertex of the parabola is now at (-4.75, -5.125), the parabola needs to be translated up by at least 5.125 units for the vertex to be placed in the second quadrant. Adding 5.125 to the equation will accomplish this.

It follows that the following equation will place the vertex in the second quadrant.

. where

Below is the graph where

Try using the slider in desmos to test this fact!

Now, let’s go back to our initial equation

The vertex of this equation is (-0.75, -5.125). The question is: What should be done to this equation to create a parabola that is concave down and has the vertex (-0.75, -5.125).

Initially, I wanted to simply negate the entire equation, but this did not provide the correct parabola.

https://www.desmos.com/calculator/oljfrj8e2m

The vertex of this equation is (-0.75, 5.125). The parabola is concave down, the x-value is correct, but the y-value is 10.25 units off. To accommodate this, -10.25 should be added to the equation which will shift the graph 10.25 units down.

The equation then becomes

https://www.desmos.com/calculator/oljfrj8e2m