SOLUTION: Oh dear God please help me!!! We are studying Graphs of Exponential and Logarithmic Functions and I am so lost I'm about to send out the hound dogs, but seriously if someone could

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Oh dear God please help me!!! We are studying Graphs of Exponential and Logarithmic Functions and I am so lost I'm about to send out the hound dogs, but seriously if someone could       Log On


   

Question 150386: Oh dear God please help me!!! We are studying Graphs of Exponential and Logarithmic Functions and I am so lost I'm about to send out the hound dogs, but seriously if someone could please help just get me started I would greatly appreciate it.
I am supposed to Plot the graphs of the following functions:I don't understand how
1. f(x) = 7x this x variable is actually an exponent so it is 7 to the x power
2. f(x) = 4x - 3 the x-3 here is also an exponent so it reads 4 to the x-3 power
3. f(x) = (1/5)x this is actually 1/5 to the x power again the x is an exponent
4. f(x) = log3x
Thank you guys for always helping me out. I can't wait till I get this degree so I can help yall out with some $$$.
Found 2 solutions by jim_thompson5910, scott8148:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
I'll do the first two to get you started.

# 1

Table of Contents:

Table
Graph

In order to graph y=7%5Ex, we need to plot a few points.


Let's find the y value when x=-2 note: you can start at any x value.


y=7%5Ex Start with the given equation.


y=7%5E-2 Plug in x=-2.



y=1%2F7%5E2 Rewrite 7%5E-2 as 1%2F7%5E2


y=1%2F49 Evaluate 7%5E2 to get 49


y=0.02 Divide and simplify.


So when x=-2, y=0.02. So we have the point (-2,0.02).


----------------------------


Let's find the y value when x=-1


y=7%5Ex Start with the given equation.


y=7%5E-1 Plug in x=-1.


y=1%2F7%5E1 Rewrite 7%5E-1 as 1%2F7%5E1


y=1%2F7 Evaluate 7%5E1 to get 7


y=0.143 Divide and simplify.


So when x=-1, y=0.143. So we have the point (-1,0.143).


----------------------------


Let's find the y value when x=0


y=7%5Ex Start with the given equation.


y=7%5E0 Plug in x=0.


y=1 Evaluate 7%5E0 to get 1


So when x=0, y=1. So we have the point (0,1).


----------------------------


Let's find the y value when x=1


y=7%5Ex Start with the given equation.


y=7%5E1 Plug in x=1.


y=7 Evaluate 7%5E1 to get 7


So when x=1, y=7. So we have the point (1,7).


----------------------------


Now let's make a table of the values we just found.





Jump to Table of Contents

Table of Values:


TEST: -2



xy
-20.02


-10.143


01


17

Now let's plot these points:







Jump to Table of Contents

Graph:


Now draw a curve through all of the points to graph y=7%5Ex:


Graph of y=7%5Ex




# 2

Table of Contents:

Table
Graph

In order to graph y=4%5E%28x-3%29, we need to plot a few points.


Let's find the y value when x=0 note: you can start at any x value.


y=4%5E%28x-3%29 Start with the given equation.


y=4%5E%280-3%29 Plug in x=0.


y=4%5E%28-3%29 Subtract


y=1%2F4%5E3 Rewrite 4%5E%28-3%29 as 1%2F4%5E3


y=1%2F64 Evaluate 4%5E3 to get 64


y=0.016 Divide and simplify.


So when x=0, y=0.016. So we have the point (0,0.016).


----------------------------


Let's find the y value when x=1


y=4%5E%28x-3%29 Start with the given equation.


y=4%5E%281-3%29 Plug in x=1.


y=4%5E%28-2%29 Subtract


y=1%2F4%5E2 Rewrite 4%5E%28-2%29 as 1%2F4%5E2


y=1%2F16 Evaluate 4%5E2 to get 16


y=0.063 Divide and simplify.


So when x=1, y=0.063. So we have the point (1,0.063).


----------------------------


Let's find the y value when x=2


y=4%5E%28x-3%29 Start with the given equation.


y=4%5E%282-3%29 Plug in x=2.


y=4%5E%28-1%29 Subtract


y=1%2F4%5E1 Rewrite 4%5E%28-1%29 as 1%2F4%5E1


y=1%2F4 Evaluate 4%5E1 to get 4


y=0.25 Divide and simplify.


So when x=2, y=0.25. So we have the point (2,0.25).


----------------------------


Let's find the y value when x=3


y=4%5E%28x-3%29 Start with the given equation.


y=4%5E%283-3%29 Plug in x=3.



y=4%5E%280%29 Subtract


y=1 Evaluate 4%5E0 to get 1


So when x=3, y=1. So we have the point (3,1).


----------------------------


Let's find the y value when x=4


y=4%5E%28x-3%29 Start with the given equation.


y=4%5E%284-3%29 Plug in x=4.


y=4%5E%281%29 Subtract


y=4 Evaluate 4%5E1 to get 4


So when x=4, y=4. So we have the point (4,4).


----------------------------


Now let's make a table of the values we just found.





Jump to Table of Contents

Table of Values:


TEST: 0



xy
00.016


10.063


20.25


31


44

Now let's plot these points:







Jump to Table of Contents

Graph:


Now draw a curve through all of the points to graph y=4%5E%28x-3%29:


Graph of y=4%5E%28x-3%29

Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
$$$, huh...WOW!!

plotting the graphs is straight forward (a graphing calculator would be a big help)
__ the f(x) values are on the vertical (y) axis and the x values are on the horizontal axis
__ find the f(x) values by "plugging in" values for x

1. f(x)=7^x __ when x=0, f(x)=1 (this is the y-intercept) __ when x is 1, f(x)=7
__ as x becomes a large NEGATIVE value, f(x) approaches zero (horizontal asymptote)

2. f(x)=4^(x-3) __ when x=3, the exponent is 0 so f(x)=1 __ when x=0, f(x)=4^(-3) or 1/64
__ same general shape as #1 with different y-intercept

3. f(x)=(1/5)^x __ when x=0, f(x)=1
__ as x becomes a large POSITIVE value, f(x) approaches zero (horizontal asymptote)
__ this graph is sort of a "mirror image" of #'s 1 and 2

4. logarithms are NOT defined for negative quantities, so this graph is only on the right-hand side of the vertical axis
__ as x approaches zero (very small fractions), f(x) approaches negative infinity (vertical asymptote)