Algebra 2 maximum height question

Question

The height of an object in motion at time "t" seconds is given by h(t)=-t^2 + 6t +10. Find the maximum height.

Answer 1

Could also use -b/2a to find vertex.

-t² + 6t +10; a = -1; b = 6

-6/2(-1) = 3 so maximum occurs at t = 3

-(3²) + 6(3) + 10 = -9 + 18 + 10 = 19

Why not plot it on Desmos Graphing Calculator to visualize the function?

Answer 2

h(t)= -t^2 +6t +10

=-(t^2-6t +9) +10 +9

= -(t-3)^2 +19 which is a downward opening parabola with vertex =maximum (3,19)=(t,h(t))

y=19 is the maximum height

or using calculus

h'(t) = -2t+6=0

t = 6/2=3

h(3)=-(3^2)+6(3)+10= -9+18+10= 19= max