The vertex is the maximum or minimum point of the equation's parabola
The general form of a quadratic equation is :
[tex]\text{ y = ax}^2\text{ + bx + c}[/tex]If we select the first two points i.e
[tex](0,\text{ 5) and (0.25, 8)}[/tex]We substitute each point to get:
[tex]5\text{ = c}[/tex][tex]\begin{gathered} 8\text{ = 0.0625a + 0.25b + 5} \\ 0.0625a\text{ + 0.25b = 3 eqn (1)} \end{gathered}[/tex]Select point (0.50, 9) and substitute into our general expression:
[tex]\begin{gathered} 9\text{ = 0.25a + 0.5b + 5} \\ 0.25a\text{ + 0.5b = 4 eqn (2)} \end{gathered}[/tex]Solving eqn (1) and (2) simultaneously, we have
[tex]\begin{gathered} a\text{ = -16} \\ b\text{ =16} \end{gathered}[/tex]The quadratic equation is given as :
[tex]H=-16t^2\text{ + 16t + 5}[/tex]The x-coordinate of the vertex can be obtained as:
[tex]\begin{gathered} =\text{ }\frac{-b}{2a} \\ =\text{ }\frac{-16}{2\text{ }\times\text{ -16}} \\ =\text{ 0.5} \end{gathered}[/tex]y-coordinate is obtained as :
[tex]\begin{gathered} H=-16(0.5)^2\text{ + 16(0.5) + 5} \\ =\text{ 9} \end{gathered}[/tex]The correct option should be the time in seconds when the ball reaches the maximum height