Question

What coordinates point can I plot in the graph? It can't be a decimals point, It has to be whole numbers because the graph doesn't let me put for example 4.86 or a fraction

Answer 1

`y=((1)/(2))^(x+1)-9`

First, we can start input small whole numbers in the function to see if we get a whole number. Let's try with 0, 1 and 2:

`\begin{gathered} y=((1)/(2))^(0+1)-9=(1)/(2)-9=-8.5 \\ . \\ y=((1)/(2))^(1+1)-9=(1)/(4)-9=-8.75 \\ . \\ y=((1)/(2))^(2+1)-9=(1)/(8)-9=-8.875 \end{gathered}`

None of those values worked. Let's try negative integers, such as -1 and -2:

`\begin{gathered} y=((1)/(2))^(-1+1)-9=((1)/(2))^0-9=1-9=-8 \\ . \\ y=((1)/(2))^(-2+1)-9=((1)/(2))^(-1)-9=2-9=-7 \end{gathered}`

We get integer coordinates if we use negative integers. Then, we need at least 5 points. Let's use x = -1, -2, -3, -4, -5

`\begin{gathered} y=((1)/(2))^(-3+1)-9=((1)/(2))^(-2)-9=2^2-9=4-9=-5 \\ . \\ y=((1)/(2))^(-4+1)-9=((1)/(2))^(-3)-9=2^3-9=8-9=-1 \\ . \\ y=((1)/(2))^(-5+1)-9=((1)/(2))^(-4)-9=2^4-9=16-9=7 \end{gathered}`

We have the points:

(-1, -8), (-2, -7), (-3, -5), (-4, -1) and (-5, 7)

If we locate them in the cartesian plane:

And now, we can estimate the graph. An exponential function where the base is smaller than 1, describes an exponential decay, and the term "-9" is applying a shift 9 units down, and also y = -9 is a horizontal asymptote. With all this and the points we just found: