Answer:
<em>Explanation below</em>
Step-by-step explanation:
<u>Vertical Shift of Functions</u>
Given a linear function
y = mx + b
The value of b is called the y-intercept or the y-coordinate where the line crosses the y-axis. If the function is shifted up by a value of k, then the new function is:
y' = mx + b + k
Similarily to shift the function down, the value of k is subtracted:
y''= mx + b - k
We are given the function:
![\displaystyle y=\frac{2}{3}x+2](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%3D%5Cfrac%7B2%7D%7B3%7Dx%2B2)
1) To shift up the line, we select a value of k=10. The new shifted-up function is:
![\displaystyle y=\frac{2}{3}x+2+10](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%3D%5Cfrac%7B2%7D%7B3%7Dx%2B2%2B10)
![\displaystyle y=\frac{2}{3}x+12](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%3D%5Cfrac%7B2%7D%7B3%7Dx%2B12)
2) To shift down the line, we select a value of k=5. The new shifted-down function is:
![\displaystyle y=\frac{2}{3}x+2-5](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%3D%5Cfrac%7B2%7D%7B3%7Dx%2B2-5)
![\displaystyle y=\frac{2}{3}x-3](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%3D%5Cfrac%7B2%7D%7B3%7Dx-3)
3) To shift to the origin, we must select a specific value of k that cancels the y-intercept. We must shift down by k=2:
![\displaystyle y=\frac{2}{3}x+2-2](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%3D%5Cfrac%7B2%7D%7B3%7Dx%2B2-2)
![\displaystyle y=\frac{2}{3}x](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%3D%5Cfrac%7B2%7D%7B3%7Dx)
This new function has the y-intercept equal to 0