Answer:
Vertex (-2.5,-300)
Step-by-step explanation:
We are given with h(t)=-16 t^2 +80 t. We are asked to find the vertex.
X-coordinate of vertex =[tex]\frac{-b}{2a}[/tex]
Where 'a' is coefficient of [tex]t^{2}[/tex] term
'b' is coefficient of 't' term.
From given function, a= -16
b=80
So,
x-coordinate of vertex =[tex]\frac{-80}{2(-16)}[/tex]
=-2.5
y-coordinate of vertex =[tex]-16(-2.5)^{2} +80(-2.5)[/tex]
Simplify it,
=-100 -200
= -300
So, vertex is at (-2.5, -300).
