Let C : x^2-4x+2k be a parabola, with vertex m. By taking derivatives or otherwise discuss, as k varies, the coordinates of m and, accordingly, the number of solutions of the equation x^2-4x+2k=0. Illustrate your work with graphs

Write y=x²-4x+2k. And m:= (x_m, y_m) for the coordinates of our vertex. We deduce that x_m  is exactly the value of x for which y'=2x-4=0, because m is a minimum point of y. By solving y'=0, we get x=2=x_m, so by plugging it into our initial equation y=x²-4x+2k we obtain y_m, which, as it should be, depends on k. That is : y_m= (2)²-4(2)+ 2k= 2k-4. So we can now write : m:= (x_m, y_m)= ( 2, 2k-4). Now we are ready to discuss x_m and y_m  as k varies, in particular we distinguish three cases: (i) y_m= 2k-4=0, from which follows k=2 and so m:= (x_m, y_m)= ( 2, 0). This means that m lies on the x-axis and that our parabola intersects the y-axis at (0, 2k)=(0,4), we sketch this and we deduce that this case corresponds to =x²-4x+2k= 0 having just one solution, namely x=2. We note that this is the same as putting b²-4ac=0, where a,b and c are coefficients of ax²+bx+c.(ii)y_m= 2k-4>0, from which follows k>2 and so m:= (x_m, y_m)= ( 2, y_m>0). This means that m lies on the first quadrant and that our parabola intersects the y-axis at (0, 2k > 4), as k>2, we sketch this and we deduce that this case corresponds to x²-4x+2k= 0 not having solutions. We note that this is the same as setting b²-4ac<0, where a,b and c are coefficients of ax²+bx+c. (iii) y_m= 2k-4<0, from which follows k<2 and so m:= (x_m, y_m)= ( 2, y_m< 0). This means that m lies on the fourth quadrant and that our parabola intersects the y-axis at (0, 2k < 4 ), as k<2, we sketch this and we deduce that this case corresponds to x²-4x+2k= 0 having two solutions. We note that this is the same as setting b²-4ac > 0, where a,b and c are coefficients of ax²+bx+c.