I need some help on #18

  A. the vertical intercept (often called the y-intercept)

  B. the slope of the line

  C. V(t) = 100t +2000

  D. V(10) = 3000, the value in year 10

  E. see attached

Step-by-step explanation:

You are told that year 0 is the year that you made the investment. Then the value in year 0 represents the value of the investment in the year you made the investment. (It is sometimes called the "initial value.")

When the function is graphed, the value in year 0 will be the value on the graph at t=0. If t=0 is the vertical axis (as it usually is), then 2000 is the vertical intercept--the point where the line meets the vertical axis.

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To find the difference between values in consecutive years, subtract one from the other:

  2100 -2000 = 100 . . . . the change in value in consecutive years

On the graph, for each increment of 1 in the value of t, the value of the investment will increase by this amount. That means the slope of the graph (rise/run) is 100/1 = 100. The difference in value in consecutive years is the slope of the graph.

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The equation of the line is ...

  V(t) = (slope) × t + (vertical intercept)

  V(t) = 100t +2000

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V(10) = 100×10 +2000 = 1000 +2000

V(10) = 3000 . . . . . the value of the investment in year 10

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see attached

b.-4 units

all you   do is count how many times you move

(5x-1)^2= (5x)^2-2(5x)×1+(1)^2=25x^2-10x+1