Explanation:
1.
since we only need the new slope, and the rest is irrelevant, this can be solved with the given information.
the shift up does not change the slope. the line is still inclined in the same way as before : -4.
now it is reflected. I have to assume it is across one of the coordinate axes, as there is no indication of another line or so that we could use as the "mirror" of the reflection.
the reflection charges the scope, and funny enough, for the slope it does not matter if we use the x- of the y-axis. the same thing happens to the slope : it flips the sign.
so, the new slope is : 4.
the new equation could be
y = 4x + 7
when reflecting across the y-axis.
the shift up by 2 simply added 2 to the function (5 + 2 = 7).
2.
luckily we need only the new y-intercept.
a less steep line has a smaller slope (in absolute value).
and the shift down by 10 simply subtracts 10 from the original function.
so, the new line could be
y = 2x - 14
as -4 - 10 = -14.
the new y-intercept (the y-value when x = 0) is -14.
3.
the slope doubled is 2×1 = 2.
shifted down by 1 means subtracting 1 from the equation.
so, we get
y = 2x - 8
anyway the new slope is 2.
4.
the slope tripled is 3×4 = 12.
shifting up by 9 means to add 9 to the function.
so, we get
y = 12x + 9
the new slope is 12.
the new y-intercept is 9.
the Sum is
4
-14
2
12
9
-------
13