[tex]y=(\frac{1}{2})^{x+1}-9[/tex]
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:
[tex]\begin{gathered} y=(\frac{1}{2})^{0+1}-9=\frac{1}{2}-9=-8.5 \\ . \\ y=(\frac{1}{2})^{1+1}-9=\frac{1}{4}-9=-8.75 \\ . \\ y=(\frac{1}{2})^{2+1}-9=\frac{1}{8}-9=-8.875 \end{gathered}[/tex]
None of those values worked. Let's try negative integers, such as -1 and -2:
[tex]\begin{gathered} y=(\frac{1}{2})^{-1+1}-9=(\frac{1}{2})^0-9=1-9=-8 \\ . \\ y=(\frac{1}{2})^{-2+1}-9=(\frac{1}{2})^{-1}-9=2-9=-7 \end{gathered}[/tex]
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
[tex]\begin{gathered} y=(\frac{1}{2})^{-3+1}-9=(\frac{1}{2})^{-2}-9=2^2-9=4-9=-5 \\ . \\ y=(\frac{1}{2})^{-4+1}-9=(\frac{1}{2})^{-3}-9=2^3-9=8-9=-1 \\ . \\ y=(\frac{1}{2})^{-5+1}-9=(\frac{1}{2})^{-4}-9=2^4-9=16-9=7 \end{gathered}[/tex]
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:
