For the first one, you can see that the line is a horizontal line, meaning that that the values of y are fixed, they are all the same. We can also see that the line passes through the point (0,2) So the y values is always going to be 2. The correct table should be the second table since it's the only one that has the y variables as 2.
For the second, an easy way of solving it, it's to find the values of y when x is equal to 0 So, f(x) = 3x + 5 f(0) = 3(0) + 5 f(0) = 0 + 5 f(0) = 5 So, the x value would be 0 and the y value would be 5 If you look at the table, you need to find the table which has a x value of 0 corresponding with a y value of 5 That table would be the first one.
