Question

How would you graph this?
My original equation= F(x) = 23(1.2)^x
A recent drop in sales has affected Stock D with the function g(x) = –6. Explain to Gordon how Stock D’s new price function, f(x) + g(x), will be created. Graph f(x) + g(x).
It would be like 1+ -6, 2+ -6 and so on you just substitute it into the equation to make any profit he would have to do more than 6.
F(x) - 6= 23( 1.2^x) - 6

Answer 1

so... what area you trying to do? f(x) + g(x) or graph \(\bf f(x) = 23(1.2)^x\) ?

Answer 2

it said to graph F(x) + G(x) and im not sure how to do that

Answer 3

so... ok.. what's f(x) and what's g(x) firstly ?

Answer 4

it is -6

Answer 5

what do you mean? hmm if you dunno what f(x) and g(x) are ..... kinda hard to add them up :)

I can see the g(x) = -6 up there, what about f(x) ? is it \(\bf f(x) = 23(1.2)^x?\)

Answer 6

I think you just substitute a number but i'm not sure, if x was one than it would be F(-5) because your subtracting 6 right?

Answer 7

if g(x) = -6 and \(f(x) = 23(1.2)^x\) then \(\bf f(x) - 6= 23( 1.2^x) - 6\) is correct

Answer 8

but im not sure how you would simplify it to be able to graph it,

Answer 9

well... you can;t simplify it really

Answer 10

Oh wait so it says just graph f(x) + g(x) so how would you graph F(x) - 6

Answer 11

what you can do is keep in mind that the graph, is just the graph of \(\bf 1.2^x\) only

and is "shrunk" by a factor of 23, and shifted downwards by 6 units

Answer 12

so it would end up being on 17 with the same rate of change?

Answer 13

\(\large {
\textit{function transformations}
\\ \quad \\

\begin{array}{llll}

\begin{array}{llll} shrink\ or\\ expand\\ by\ {\color{purple}{ A}}\cdot {\color{blue}{ B}}\end{array}
\qquad
\begin{array}{llll} vertical\\ shift\\ by \ {\color{green}{ D}} \end{array}

\begin{array}{llll}{\color{green}{ D}} > 0& Upwards
\\ \quad \\
{\color{green}{ D}} < 0 & Downwards\end{array}
\\

\qquad \downarrow\qquad\qquad\quad\ \downarrow\\
% template start
y = {\color{purple}{ 23}} ( 1.2^{{\color{blue}{ B}}x + {\color{red}{ C}}} ) - {\color{green}{ D}}\\
% template ends
\qquad\qquad\quad\ \uparrow \\

\qquad\begin{array}{llll} horizontal\\ shift\\ by \ \frac{{\color{red}{ C}}}{{\color{blue}{ B}}}\end{array}

\begin{array}{llll}\frac{{\color{red}{ C}}}{{\color{blue}{ B}}} > 0 & to\ the\ left
\\ \quad \\
\frac{{\color{red}{ C}}}{{\color{blue}{ B}}} < 0& to\ the\ right\end{array}
\end{array}
}\)

Answer 14

i think its supposed to be a verticle shift and just move down 6 to make the principle 16?

Answer 15

so yes

Answer 16

Thanks

Answer 17

yw