Mth 163 040203

If f(t) = .02 t^2 – 3 t + 80 then give expressions for the following:

• f(t+10)

f(t+10) =

.02 (t+10)^2 - 3 ( t + 10) + 80 =

.02 t^2 + .4 t + 2 - 3 t - 30 + 80 =

.02 t^2 - 2.6 t + 52.

• f(t+10) – f(t)

f(t+10) - f(t) =

.02 (t+10)^2 - 3 ( t + 10) + 80 - (.02 t^2 - 3 t + 80) = .02 t^2 - 2.6 t + 52 - .02 t^2 + 3 t - 80 = .4 t - 28

Notes:

f(t+10) is the y value at the t+10 point, and f(t) is the y value at the t point.

f(t+10) - f(t) is the change in the y value from the t point to the t+10 point

f(t+10) - f(t) is the change in the y coordinate of the graph from the t point to the t+10 point, which is the rise from the first to the second point.

• [ f(t + 10) – f(t) ] / 10

[f(t+10) - f(t) ] / 10 =

[ .02 (t+10)^2 - 3 ( t + 10) + 80 - (.02 t^2 - 3 t + 80) ] / 10 =

(.4 t - 28) / 10 = .04 t - 2.8

Note:

[f(t+10) - f(t) ]  is the rise from the t point to the t + 10 point

10 is the run from the t point to the t + 10 point.

So [f(t+10) - f(t) ] / 10 is rise / run = average slope between the t point and the t + 10 point.

• f(t + `dt)

f(t+`dt) =

.02 (t+`dt)^2 - 3 ( t + `dt) + 80 =

.02 t^2 + .04 t `dt + .02 `dt^2 - 3 t - 3 `dt + 80

• f(t + `dt) – f(t)

 f(t+`dt) - f(t) =

.02 (t+`dt)^2 - 3 ( t + `dt) + 80 - (.02 t^2 - 3 t + 80) =

02 t^2 + .04 t `dt + .02 `dt^2 - 3 t - 3 `dt + 80 - .02 t^2 + 3 t - 80 =

.04 t `dt - 3 `dt + .02 `dt^2

f(t+`dt) - f(t) represents the rise from the t point to the t + `dt point.

• [ f(t+`dt) – f(t) ] / `dt

[ f(t+`dt) – f(t) ] / `dt = ... = (.04 t `dt - 3 `dt + .02 `dt^2) / `dt = .04 t - 3 + .02 `dt.

f(t+`dt) - f(t) represents the rise from the t point to the t + `dt point and `dt represents the run, so [ f(t+`dt) – f(t) ] / `dt represents rise / run = average slope between the t and t + `dt points.

Very quickly make tables and construct graphs for y = t, y = t^2 amd y = 2^t.

If g(t) = t^2 then give the t = -1, t = 0 and t = 1 points of the following, and graph the points:

• g(t)

The points are (-1, 1), (0, 0) and (1, 1).

• 2 g(t)

All y values are double from before, yielding points (-1, 2), (0, 0) and (1, 2).

• .1 g(t)

All y values are .1 times as great as those for the basic function, yielding points (-1, .1), (0, 0) and (1, .1).

• g(t) + 4

All y values are 4 units higher than for the basic function, yielding points (-1, 5), (0, 4) and (1, 5).

• 2 g(t) + 4

The y values of the basic function are first doubled, yielding points (-1, 2), (0, 0) and (1, 2).  The points are then raised 4 units to give us (-1, 6), (0, 4) and (1, 6).

• .1 g(t) + 4

The y values of the basic function first become .1 as great, yielding points (-1, .1), (0, 0) and (1, .1).  The points are then raised 4 units to give us (-1, 4.1), (0, 4) and (1, 4.1).