We have to look at the 4^x term
if x> 0 , then 4^x is obviously positive.
if x= 0 then 4^0 = 1
for x < 0, we get 1/4^x which is 1/(positive number)
now for huge negative values of x, the denominator in
1/4^x becomes huge and 1/4^x approaches, but will never reach zero
so we have y > - 5 as the range.
How do you find the range of this problem?
y = 4^x - 5
Where do I even look for the range?
6 answers
So is that the "K" I'm looking at to solve for the range-my teacher keeps talking about the a h, k, ect and what effect they have
Like if it said y = 2^x + 8
the range would be y>6
Thank you
Like if it said y = 2^x + 8
the range would be y>6
Thank you
in your second example the range would be
y ≥ 8 , not 6
since 2^x > 0 for all x,
you are adding a positive value to 8, so you are looking at something > 8
y ≥ 8 , not 6
since 2^x > 0 for all x,
you are adding a positive value to 8, so you are looking at something > 8
Okay, let me try this again-I meant 8, not 6-a typo but then why is y>=8 in this one and the other one it is y>-5 and not greater and equal-is it because of the positive sign?
So for y = 3^x - 4, the range would be y >-4 and not y >= (-4), right-I'm just trying to really understand
So for y = 3^x - 4, the range would be y >-4 and not y >= (-4), right-I'm just trying to really understand
You are right, it should be y > 8 , my typo.
and your last example is correct too.
It has to be > and not ≥ since the term 3^x can never really become zero, so you always adding something which is positive.
and your last example is correct too.
It has to be > and not ≥ since the term 3^x can never really become zero, so you always adding something which is positive.
Thanks-much appreciated