1. If we know that (3, -4) is a data point on a line with slope 2.8, what is the equation of the line?

Using y = m x + b, we know that the slope is 2.8 so we have

y = 2.8 x + b.

We also know that (3, -4) is a data point so

-4 = 2.8 * 3 + b.

Solving for b we get

b = -4 - 2.8 * 3 = -12.4.

2. Find the linear function depth(t) = mt + b if it is known that at clock time 30 seconds the depth is 50 centimeters, and that depth is changing at a rate of -.35 centimeters/second.

We know that depth(30) = 50 cm and that the rate of depth change is -.35 cm/sec.  So we know depth(30), and we know m.

Since m = -.35 cm/s,

depth(t) = -.35 cm/s * t + b.

Since depth(30) = 50 cm we have

50 cm = -.35 cm/s * 30 sec + b so that

b = 50 cm -.35 cm/s * 30 sec = 50 cm - 10.5 cm = 39.5 cm.

5. Suppose a line passes through (3, -2) with slope 5. What is the slope = slope equation (y - y1) / (x - x1) = m when we substitute these values? What do we get when we solve for y?

The slope is (y - y1) / (x - x1) = (y - (-2) ) / ( x - 3) = (y + 2) / (x - 3).

The slope is 5.

So

(y + 2) / ( x - 3) = 5.

Solving for y we multiply both sides by (x - 3) to get

(y + 2) = 5 ( x - 3) so that

y + 2 = 5 x - 15 and

y = 5x - 15 - 2 or

y = 5x - 17.

Substitute the above values x = 3, y = -2 and m = 5 into the form y = mx + b. Solve for b. Then substitute m and b into the form y = mx + b to obtain the equation of the line.

Substituting we get

-2 = 5 * 3 + b so that

b = -2 - 5 * 3 = -17.

Thus our equation is

y = 5 x - 17.

Compare the equations you obtained from the different methods.

Make a statement about whether or not the two methods seem to give the same results.