Suppose the path of the ball in above figure is f(x) = -0.25(x-1)2 + 6.25. Find the ball’s initial and maximum heights.


For a quadratic function is in standard form, f(x)=ax2+bx+c.
The standard quadratic form is ax2+bx+c=y, In this the term c gives us the y-intercept of the curve.This is the point at which ball’s initial height.
The vertex form of a quadratic equation is y=a(x−h)2+k. This gives the vertex (h, k) in which y coordinate is the top most point of the ball.

The correct answer is: C = 6

    The given equation is f(x) = -0.25(x-1)2 + 6.25
    This is in the vertex form
    Therefore on comparing with the vertex form of a quadratic equation is y=a(x−h)2+k. we get,                                                                                 h = 1 , k = 6.25
    Therefore, vertex is (h,k) = (1, 6.25)
    So the maximum height of the ball is 6.25 .
    Converting the equation in standard form

    f(x) = -0.25(x-1)2 + 6.25

             f(x) = -0.25(x2+1 – 2x) + 6.25

                    f(x) = -0.25x2 – 0.25 + 0.5x + 6.25

         f(x) = -0.25x2 +0.5 x + 6.00
    Therefore, y intercept is c = 6
    Which is the initial height of the ball i.e 6