Table of value for y=2-3(5^x+4), plus its transformation from y=5^x.

[calcdummy]

Homework Statement

Describe the transformations that must be applied to the graph of y=5^x to obtain the graph of y=2-3(5^x+4) and complete the following table.

y=5^x
(-1,1/5)
(0,1)
(1,5)
(2,25)
(3,125)

y=-3(5^x)


y=2-3(5^x+4)

Homework Equations

y=5^x , y=2-3(5^x+4)

The Attempt at a Solution

y=5^x
(-1,1/5)
(0,1)
(1,5)
(2,25)
(3,125)


y=-3(5^x)
(-1,-3/5)
(0,-3)
(1,-15)
(2,-75)
(3,-375)

y=2-3(5^x+4)
(-5,3/5)
(-4,3)
(-3,15)
(-2,75)
(-1,375)

What I did to achieve the points in the second column was to multiply the y-coordinates in the first column by -3. Then I proceeded to adding "-4" to the x coordinates and multiplying the y coordinates by -1 because of "2-3".

I couldn't check to see if my points are correct on my graphing calculator because I didn't know how to input y=2-3(5^x+4). Anyway on the the transformations which I presumed to be:

Shifted to the left by 4(d)
Reflected in x-axis (because I figured "a" would end up being -1)
After that I'm stumped.


What I think the parameters are: a=-1 k=0 d=4 c=0

 

[HallsofIvy]

Homework Statement

Describe the transformations that must be applied to the graph of y=5^x to obtain the graph of y=2-3(5^x+4) and complete the following table.

y=5^x
(-1,1/5)
(0,1)
(1,5)
(2,25)
(3,125)

y=-3(5^x) y=2-3(5^x+4)

Homework Equations

y=5^x , y=2-3(5^x+4)

The Attempt at a Solution

y=5^x
(-1,1/5)
(0,1)
(1,5)
(2,25)
(3,125) y=-3(5^x)
(-1,-3/5)
(0,-3)
(1,-15)
(2,-75)
(3,-375)

y=2-3(5^x+4)
(-5,3/5)
(-4,3)
(-3,15)
(-2,75)
(-1,375)

No. "-3A" is not "2-3A". Given that when x= -1, -3(5^x) is -3/5, then 2- 3(5^x) is 2- 3/5= 2/5. You had already accounted for the "-" when you multiplied by -3 to get the second column. The new thing in the third column is the "2". Get your third column by adding 2 to the second column.

What I did to achieve the points in the second column was to multiply the y-coordinates in the first column by -3. Then I proceeded to adding "-4" to the x coordinates and multiplying the y coordinates by -1 because of "2-3".

I couldn't check to see if my points are correct on my graphing calculator because I didn't know how to input y=2-3(5^x+4). Anyway on the the transformations which I presumed to be:

Shifted to the left by 4(d)
Reflected in x-axis (because I figured "a" would end up being -1)
After that I'm stumped. What I think the parameters are: a=-1 k=0 d=4 c=0

 

[calcdummy]

Are you saying that everything prior to the third column is correct and that I should just add "2" to both the x and y-coordinates?

 

[calcdummy]

I'm sorry but could someone please help me out... HallsofIvy's comment left me a bit confused.

 

[physicsl0ver]

I'm sorry but could someone please help me out... HallsofIvy's comment left me a bit confused.

y=2-3(5^x+4) solved is y=-3(5^x)-10

Try using that equation in the last column. Should just have an increment of -10 in the y's.

You can sub y=2-3(5^x+4) in wolfram alpha (online) to see if the solved equation satisfies the points for the last column.
Lemme know if that helps at all or if you have any questions.

Best of luck

 

[physicsl0ver]

No. "-3A" is not "2-3A". Given that when x= -1, -3(5^x) is -3/5, then 2- 3(5^x) is 2- 3/5= 2/5. You had already accounted for the "-" when you multiplied by -3 to get the second column. The new thing in the third column is the "2". Get your third column by adding 2 to the second column.

Not quite understanding your math.

2- 3/5 = 2/5?

2- 3/5 = 7/5 = 1 2/5

so you can write it as either 7/5 or 1 2/5.

 

[Joffan]

Here's how I'd show the step-by-step transform:
$$\begin{array} \\<br /> x & 5^x & 5^x+4 & 3(5^x+4) & 2-3(5^x+4) \\<br /> -1 & 0.2 & 4.2 & 12.6 & -10.6 \\<br /> 0 & 1 & 5 & 15 & -13 \\<br /> 1 & 5 & 9 & 27 & -25 \end{array}$$
Unless you're required to reduce this to one multiplication and one addition, I don't really understand why you're asked to show $-3(5^x)$.

Note - in support of HallsofIvy's point - that 2-3A = -3A + 2.