Question

The topic of the assignment is: Graphing Decay Functions I will post a screenshot of the function that I need to graph. The function follows this formula: f(x)=a(b)^(x-h)+k

Answer 1

We are asked to graph the following function:

$y=\frac{1}{2}(\frac{1}{6})^{x-1}-2$

Before sketching the graph we will simplify the function. To do that we will use the following property of exponentials:

$a^{x+y}=a^xa^y$

Applying the property we get:

$y=\frac{1}{2}(\frac{1}{6})^x(\frac{1}{6})^{-1}-2$

Now we use the following property:

$(\frac{a}{b})^{-1}=\frac{b}{a}$

Applying the property we get:

$y=\frac{1}{2}(\frac{1}{6})^x(6)-2$

Simplifying we get:

$y=3(\frac{1}{6})^x-2$

Now, we will graph the function. To do that we will need to determine three points of the graph.

The first point we will determine it by setting x = 0, to determine the y-intercept:

$\begin{gathered} y=3(\frac{1}{6})^0-2 \\ \\ y=3-2=1 \end{gathered}$

Therefore, the point:

$(x,y)=(0,1)$

is part of the graph. Now, we will set x = 1:

$y=3(\frac{1}{6})^1-2$

Solving the operations:

$y=-\frac{3}{2}$

Therefore, the point:

$(x,y)=(1,-\frac{3}{2})$

Now, we will determine the horizontal asymptote. To do that we need to analyze what happens when x goes to infinity. As the values of "x" get larger and larger the value of:

$(\frac{1}{6})^x$

will tend to zero, since would have a large number of products of 1/6 and as the denominator grows the numerator will stay as 1 and therefore, the fraction will go to zero, therefore, the asymptote is:

$y=3(0)-2=-2$

Now, we plot the points and the asymptote, we get the following graph:

The direction of the graph is due to the fact that the function is a decay function.