I need help with graphing exponential functions!?

Question

I really don't know how to ask this because it's a multiple choice but could you describe the graphs [graphing calculator was STOLEN] for each of these?????!!!!PLEASE HELP ASAP!!!

[A] y=1.1(3)^x
[B] y=1.1(0.1)^x
[C] y=3(1.10^x
[D] y=((3)(1.1))^x

if you don't know already the "^" symbol is raising that number to that power following the "^" symbol.


OR you could help me learn how to do these by hand!!!

Answer 1

I'll go into in-depth detail on describing exponential curves by just looking at their equation. I believe you will find schoolwork MUCH easier once you understand how it works, but if you don't have the time to read all of that now, my answer to your original question is at the bottom.

Here we go!

Exponential functions can be generalised as y=a(b)^x where 'a' and 'b' are constants. The function means ['a' multiplied by ('b' to the power of 'x')], and can be written in the following ways:

y=a(b)^x
y=a(b^x)
y=a * b^x
y=a * (b^x)

All of the above are exactly the same, it's just that the paranthesis () are in different places. Also, a(b) means 'a' times 'b'


'a' is the y-intercept. If you want to understand why, consider this: ANY number raised to the power of 0 equals 1. (e.g. 5^0=1, 10^0=1, 5.25453^0=1) Combine this with the fact that the y-intercept is found when x=0, and look at [A] and [B]:

[A]
When x=0,
y=1.1 * 3^0
=1.1 * 1
=1.1

[B]
When x=0,
y=1.1 * 0.1^0
=1.1 * 1
=1.1


Moving on... back to y=a(b)^x, the size of 'b' determines how 'steep' your exponential curve is. Compare y=1.1(3)^x with y=1.1(2)^x:


[y=1.1(3)^x]
When x=1, y=1.1 * 3^1
When x=2, y=1.1 * 3^2
When x=3, y=1.1 * 3^3

When x=1, y=1.1 * 3
When x=2, y=1.1 * 3 * 3
When x=3, y=1.1 * 3 * 3 * 3

When x=1, y=3.3
When x=2, y=9.9
When x=3, y=29.7


[y=1.1(2)^x]
When x=1, y=1.1 * 2^1
When x=2, y=1.1 * 2^2
When x=3, y=1.1 * 2^3

When x=1, y=1.1 * 2
When x=2, y=1.1 * 2 * 2
When x=3, y=1.1 * 2 * 2 * 2

When x=1, y=2.2
When x=2, y=4.4
When x=3, y=8.8

As you can see, the curve 'grows' much more quickly for b=3 than for b=2


Special note: If 'b' EQUALS 1, then the function becomes a HORIZONTAL LINE (i.e. it's not an exponential function anymore). If 'b' is a POSITIVE number which is LESS THAN 1, then your graph will CURVE DOWNWARDS.

3(1)^1 = 3 * 1 = 3
3(1)^2 = 3 * 1 * 1 = 3
3(1)^3 = 3 * 1 * 1 * 1 = 3

1.1(0.1)^1 = 1.1 * 0.1 = 0.11
1.1(0.1)^2 = 1.1 * 0.1 * 0.1 = 0.011
1.1(0.1)^3 = 1.1 * 0.1 * 0.1 * 0.1 = 0.0011

########################################
##FINALLY, TO DESCRIBE YOUR GRAPHS...##
########################################
[A] A 'normal' ('upwards') exponential curve with y-intercept at 1.1
[B] A 'downwards' exponential curve with y-intercept at 1.1
[C] A 'normal' exponential curve. Its y-intercept is higher than [A], but it's curve is less 'steep'.

[D] Ah, a trick question...
y = ((3)(1.1))^x
= (3 * 1.1)^x
= 3.3^x
= 1(3.3)^x

...so [D] is a 'normal' exponential curve with y-intercept at 1 and a curve that's even steeper than [A].

Hope this helps, good luck!

Answer 2

definite, I consider the votes that this is a foul question. Logarithms and exponentials are inverse purposes. Graph them via reflecting them over the graph of y=x. Ask you instructor for extra rationalization, too. Your instructor will tremendously much easily have the prefer to make the relaionships between logarithmic and exponential purposes extra sparkling to you than your modern awareness. sturdy success! :)

Answer 3

i can do you one better.

if you go to http://www.calculator.com/calcs/GCalc.html

type in

1.1 * 3^x

1.1 * .1^x

3 * 1.1^x

(3.3)^x, by the way, (3 * 1.1) = 3.3

Click enter on the keyboard for each one, and leave out the y = part.