Question 428552
Solution:Since the coefficient beside {{{x^2}}}, is negative we have a downward parabola.
To find x-intercepts we solve the equation:{{{-x^2+4x=0}}} => x(4-x)=0 => x=0 and x=4
Therefore the x-intercepts are:(0,0) and (4,0)

The vertex of parabola is the point (-b/2a, f(-b/2a),substitute:

x=-4/-2=2 and {{{y=-(2)^2+4*2}}}, y=4 Thus the vertex is: (2,4)

{{{graph(300, 300, -5, 5, -5, 5, -x^2+4x)}}}.