Find the quadratic function which contains the points (0,5), (1,2) and (3,2). Graph the function

Answer 1

since quadratics are symmetric, the vertex of this parabola occurs when x = 2.

The y-intercept is at y=5, so

y = a(x-2)^2 + k
y(0) = 4a+k = 5
y(1) = a+k = 2

solve for a and k, and you have

y = (x-2)^2 + 1

verify at

http://www.wolframalpha.com/input/?i=plot+y%3D%28x-2%29^2+%2B+1%2C+y%3D2+for+x%3D0..4