We are asked to find the midpoint of the segment joining the intersection points of the curve `y=x^2-4x+4` and the line `y=x+c` (we are told that m is a constant and m=1—we let c vary.) ****If you really meant `y=x^2-4+4` (which means`y=x^2`) the following steps would still apply.****

A line and a parabola (the graph of a quadratic function) will either not intersect at all, intersect at exactly one point (the line is tangent to the curve), or at at most two points which we will label A and B. Let the midpoint be M.

To find the intersection of two curves we set the equations equal to one another (essentially using the substitution method, substituting for y.) Thus:

`x^2-4x+4=x+c` or `x^2-5x+(4-c)=0`

We can use the quadratic formula (or completing the square) to find the x-coordinates of the intersections.

`x=(5 +- sqrt(25-4(1)(4-c)))/2` or `x=(5+-sqrt(9+4c))/2`

Thus the x-coordinates of the intersections depend on the value of c that is chosen. Also note that if c<-2.25 we no longer have real solutions, so for c<-2.25 the line will not intersect the curve. Also, for c=-2.25 the line intersects the curve at one point.

To find the y-coordinates we could substitute the x-coordinates into the equation for the parabola, but we realize the intersection points also lie on the line so the y-coordinates are the x-coordinates plus c.

To find the midpoints we take the average of the x-coordinates of the intersections to get the x-coordinate of the midpoint.

`((5+sqrt(9+4c))/2+(5-sqrt(9+4c))/2)/2=5/2`

The y-coordinate of the midpoints lies on the line y=x+c so the coordinates for the midpoint of `bar(AB)` for a given choice of c is `(5/2,5/2+c)`

See the attachment for some choices of c.

**Further Reading**

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