Answer:
The maximum height of the basketball is of 24.25 feet.
Step-by-step explanation:
Vertex of a quadratic function:
Suppose we have a quadratic function in the following format:
[tex]f(x) = ax^{2} + bx + c[/tex]
It's vertex is the point [tex](x_{v}, y_{v})[/tex]
In which
[tex]x_{v} = -\frac{b}{2a}[/tex]
[tex]y_{v} = -\frac{\Delta}{4a}[/tex]
Where
[tex]\Delta = b^2-4ac[/tex]
If a<0, the vertex is a maximum point, that is, the maximum value happens at [tex]x_{v}[/tex], and it's value is [tex]y_{v}[/tex].
Height of the basketball:
Given by the following function:
[tex]h(t) = -4t^2 + 10t + 18[/tex]
Which is a quadratic function with [tex]a = -4, b = 10, c = 18[/tex]
What is the maximum height of the basketball?
y(in this case h) of the vertex. So
[tex]\Delta = b^2-4ac = 10^2 - 4(-4)(18) = 388[/tex]
[tex]y_{v} = -\frac{388}{4(-4)} = 24.25[/tex]
The maximum height of the basketball is of 24.25 feet.
