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:= (xm, ym) for the coordinates of our vertex. We deduce that xm 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=xm, so by plugging it into our initial equation y=x2-4x+2k we obtain ym, which, as it should be, depends on k. That is : ym= (2)2-4(2)+ 2k= 2k-4. So we can now write : m:= (xm, ym)= ( 2, 2k-4). Now we are ready to discuss xm and ym as k varies, in particular we distinguish three cases: (i) ym= 2k-4=0, from which follows k=2 and so m:= (xm, ym)= ( 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)ym= 2k-4>0, from which follows k>2 and so m:= (xm, ym)= ( 2, ym>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) ym= 2k-4<0, from which follows k<2 and so m:= (xm, ym)= ( 2, ym< 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.
