How can I graph the trajectory of an object that is launched from the surface of the Earth at escape velocity?

What you're looking for is a differential equations class. You know that F=ma and F=GMm/r^2, so ma=GMm/r^2, or

a = GM/r^2.

But a is the second derivative of r, so

r''(t) = GM/r^2.

This is a differential equation that can be solved using the techniques of an ODE course.

I have been toying with this same question.

I came up with a function that has what I think is the correct profile, it is just not positioned properly in the coordinate plane to make proper sense.

The function in simplest form is

y = 11.2x - (199218.9/(6400+y)^2) x^2

Compare this with the standard function of an object moving through a constant gravitational field

y = 11.2x - .0049x^2

The function values are converted to km to keep the scale of the graph smaller.

In the first equation a is expressed as a function of the radius of the earth, where y = r and r of the earth is 6400km.

so 1/2 GM/r^2 = (199218.9/(6400+y)^2)

where r = 6400+y

This assumes the starting point of the calculations to begin at the surface of the earth. So at the start y = 0 km above the surface.

The meaningful part of the graph begins below the x axis at

y = -12800 . This is where the 2 functions intersect.

If one were to reposition the graph at this point to the origin,

This would accurately reflect the position in time of the moving object. You have to disregard the part of the function that resembles a hyperbola opening up and to the right.

Obviously the parameters of the function require massaging to reposition it and eliminate what looks like the negative half of the function. Similar to how y = +√(25 - x^2) retains the positive half of a circle.