Question

h(t)=1/[(t^2+6t)(t+2)] find the domain and range, i get the domain but how do u find the range?

Answer 1

find the points that h(t) cannot b equal to and say everywhere but there

Answer 2

look for horizontal asymptotes.

Answer 3

let y be h(t)...k......now make t function of y

Answer 4

range is all values that are produced from teh domain

Answer 5

usually defined as the max or min or some such of the function

Answer 6

@amis is there a specific formula i should be going by tho?

Answer 7

If you multiply the denominator out you get
\[h(t) = \frac{1}{t^3+8t^2+12t}\]
Think about what a cubic polynomial looks like. This one goes from -infinity to +infinity, taking on negative and positive values. However, the only way to make a fraction = 0 is for the numerator to be = 0, but 1 will never = 0, so the only number you will never get for h(t) is 0. Thus the range is
\[(-\infty, 0) \cup (0,\infty)\]

Answer 8

and that is what confuses me because the range for this one is (-infi,-2) union (0,infi)

Answer 9

The domain is all real numbers except-6, -2, and 0 right?

Answer 10

specific formula ... not that i can recall

Answer 11

its always a good idea to graph the equation and so you can see its movements

Answer 12

yes the domain is -6 -2 0