How to find the range of f(t)=t^2-6t

The range is the highest y value to the lowest y value

ok

find the vertex because it is the highest or lowest point in the equation
make into f(t)=a(x-h)²+k form
(h,k) is the vertex
k is the y value of the vertex
if a>0, then the vertex is the minimum
if a<0, then the vertex is the maximum

so
f(t)=t²-6t
f(t)=t²-6t+9-9
f(t)=(t-3)²-9
f(t)=1(t-3)-9

k=-9
a=1
1>0
so y=-9 is the minimum value
the maximum value will be infinity

so the range is from y=-9 to infintiy
so R=[9,∞)

Hvhvichcufxif ufxu☺️☺️