# College Algebra: Exponential Functions How do you determine the vertical/horizontal stretch and/or vertical/horizontal shift of a function such as: f(x)=14^-x? The textbook and online instructor...

College Algebra: Exponential Functions

How do you determine the vertical/horizontal stretch and/or vertical/horizontal shift of a function such as: f(x)=14^-x?

The textbook and online instructor video only gives examples such as f(x)=2^(x+5)-3, so we know that there is a shift of 5 units to the LEFT and a shift of 3 units down from the original f(x)=2^x. I also know to make an x/y chart of values and plot the numbers, but the online textbook/exercises will not let you put any point on the graph that doesn't fit within a [-10,10] by [-10,10] window.

I'm lost on questions I keep missing like the f(x)=14^-x, and f(x)=4^-x, and f(x)=10^-x.

*print*Print*list*Cite

### 2 Answers

If `f(x) = b^x` which is the standard exponential equation has a negative x, it effectively inverts the graph. Note how `bgt1` and remember that negative exponents indicate fractions. If always simplifies matters if you substitute values in to x which then gives you co-ordinates, allowing you to plot any graph at all. Use your x/y chart of values to help you. You can usually use the x=-1, x=0 and x=1

Let's start by looking at the function `f(x)=14^(-x)` (in black)and compare it to the graph of `g(x)=14^x ` (in red).

Then look at the following graph and the green line representing `h(x)= 14^(-x+1)` and you will note that `f(x)=14^(-x)` has shifted in a horizontal direction towards the right.

To understand this, substitute values for x in the equation of `f(x)=14^(-x)` and `h(x) = 14^(-x+1)` using your chart of values. In other words, you can plot any of your graphs using your x/y table or chart. Just use the values of x that work for what you have.

Therefore, for the first equation of `f(x) = 14^(-x)` ,if x=0, and remembering the special case of zero as the exponent where anything to the power of zero is 1; for example, `b^0=1` even a randomly selected number, say, `4^0=1` , and in this case `f(x)=14^0=1` when x=0, we have the first co-ordinate (0;1) and making x=- 1, we get `f(x) = 14^(-(-1))=14` so we have co-ordinate (-1;14)and for x=1, `f(x)=14^(-(1))` which`= 1/14` we have co-ordinate (1;1/14)

Note, in the case of the h(x) equation, x=-1 would give us a large number, off the graph, `` ,so we do not need to use it.

Ah I just finished my homework in this! haha

With exponentials they only have horizontal asymptotes which is basically the x=? line in the exponential. When exponentials do not have a shift from up or down the asymptote is most likely zero. For example f(x)=14^-x would have an horizontal asymptote of 0 where as f(x)=2^(x+5)-3 would have a horizontal asymptote of -3.

The reason I guess you keep getting these, f(x)=14^-x, and f(x)=4^-x, and f(x)=10^-x, questions wrong is because first you have to graph at least two points as the orginal equations f(x)=14^x f(x)=4^x, and f(x)=10^x. Then reflect the points over the y-axis

graph1 shows14^x

graph2 shows 14^-x

**Images:**