# Show that `h(x) = 36(1 - 6^x)` is a decay function.

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### 2 Answers

`h(x)=36(1-6^x)`

Take note that a decay function is a decreasing function. It means as the value of x increases, the value of y decreases.

So to determine if it is a decay function, plug-in values to x and solve for the corresponding value of y.

`x=-2`

`y=36(1-6^(-2))=35`

`x=-1`

`y=36(1-6^(-1))=30`

`x=0`

`y =36(1-6^0)=0`

`x=1`

`y=36(1-6^1)=-180`

`x=2`

`y=36(1-6^2)=-180-1260`

`x=10`

`y=36(1-6^(-10))=-2176782300`

So we have:

x | -2 -1 0 1 2 10

y | 35 30 0 -180 -1260 -2176782300

Notice that as the values of x increases the values of y decreases.

**Hence, `h(x) = 36(1-6^x)` is a decay function.**

A decay function is when x-values increase and y-values decrease. You can prove that `h(x)=36(1-6^(x))` is a decay function by simply plugging in numbers for x. You really don't need to do more than 3 values to see the trend. I'll start with 1.

`h(x)=36(1-6^(1))=-180 `. So you know when x=1, y=-180.

`h(x)=36(1-6^(2))=-1260 `. When x=2, y=-1260. The y value decreased.

`h(x)=36(1-6^(3))=-7740 `. When x=3, y=-7740.

By now you can tell that as x increases, y decreases which fits the definition of a decaying function.

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