Answer:
<u>Linear relationship</u>: increasing or decreasing one variable will cause a corresponding increase or decrease in the other variable.
<u>Inverse relationship</u>: the value of one variable decreases as the value of the other variable increases.
<u>Exponential relationship</u>: a constant change in the independent variable (x) gives the same proportional change in the dependent variable (y)
<u>Question 7</u>
As the x-value increases (by one unit), the y-value decreases.
Therefore, this is an inverse relationship.
The y-values are calculated by dividing 16 by the x-value.
![\sf y=\dfrac{16}{x}](https://tex.z-dn.net/?f=%5Csf%20y%3D%5Cdfrac%7B16%7D%7Bx%7D)
<u>Question 8</u>
As the x-value increases, the y-value increases.
The y-value increases by a factor of 2 for each x-value increase of 1 unit from 1 ≤ x ≤ 4 and 5 ≤ x ≤ 6.
![\sf y=2^x \ for \ 1\leq x\leq 4](https://tex.z-dn.net/?f=%5Csf%20y%3D2%5Ex%20%5C%20for%20%5C%201%5Cleq%20x%5Cleq%204)
![\sf y=2^{(x+3)} \ for \ 5\leq x \leq 6](https://tex.z-dn.net/?f=%5Csf%20y%3D2%5E%7B%28x%2B3%29%7D%20%5C%20for%20%5C%205%5Cleq%20x%20%5Cleq%206)
These are separate exponential relations for restricted domains.
So there doesn't appear to be one relationship for a non-restricted domain.