Points, Vectors, and Functions Exercises

Answer

• Going from left to right we put t = 0 at (4, 8) and t = 1 at (10, 4).

First look at x and t. When t = 0 we have x = 4 and when t = 1 we have x = 10.

The intercept is 4 and the slope is 6, therefore

x(t) = 4 + 6t.

Now look at y and t. When t = 0 we  have y = 8 and when t = 1, y = 4.

The intercept is 8 and the slope is -4, therefore

y = 8 – 4t.

• Going from right to left we put t = 0 at (10, 4) and t = 1 at (4, 8).

First look at x and t. When t = 0 we have x = 10 and when t = 1 we have x = 4.

The intercept is 10 and the slope is -6, therefore

x(t) = 10 – 6t.

Now look at y and t. When t = 0 we have y = 4 and when t = 1 we have y = 8.

The intercept is 4 and the slope is 4, therefore

y(t) = 4 + 4t.

Answer

Answer

To parametrize the line that takes 4 time steps to travel from (0, 5) to (20, 1), put t = 0 at (0, 5) and t = 4 at (20, 1).

Look at x and t. When t = 0 we have x = 0 and when t = 4 we have x = 20. 

The intercept is 0 and the slope is  ,

so the equation for x becomes x = 0 + 5t = 5t. Now look at y and t. When t = 0 we have y = 5 and when t = 4 we have y = 1.

The intercept is 5 and the slope is

The equation for y is y = 5 – 1t = 5 – t.

Answer

To parametrize the line that takes 0.5 time steps to travel from (0, 5) to (20, 1), put t = 0 at (0, 5) and t = 0.5 at (20, 1).

Look at x and t. When t = 0 we have x = 0 and when t = 0.5 we have x = 20.

The intercept is 0 and the slope is the equation for x is x(t) = 40t.

Now look at y and t.

When t = 0 we have y = 5 and when t = 0.5 we have y = 1.

The intercept is 5 and the slope is so the equation for y is

y(t) = 5 – 8t.

Answer

If x = 5t then

 .

Therefore

Answer

If x = 3t – 2, then

so

.

Answer

If x = 4t + 1 then

so

.