SOLUTION: Hi need some help with this one: Find the vertex, graph and find the x-intercepts of the parabola: y=-x^2+4x Thanks!

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Hi need some help with this one: Find the vertex, graph and find the x-intercepts of the parabola: y=-x^2+4x Thanks!      Log On


   

Question 428552: Hi need some help with this one:
Find the vertex, graph and find the x-intercepts of the parabola:
y=-x^2+4x
Thanks!
Answer by Gogonati(855) About Me  (Show Source):
You can put this solution on YOUR website!
Solution:Since the coefficient beside x%5E2, is negative we have a downward parabola.
To find x-intercepts we solve the equation:-x%5E2%2B4x=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=-%282%29%5E2%2B4%2A2, y=4 Thus the vertex is: (2,4)
graph%28300%2C+300%2C+-5%2C+5%2C+-5%2C+5%2C+-x%5E2%2B4x%29.