**Graph several sets of curves for**

**for selected values of a, b, and k in an appropriate range
for t.**

We can start by setting the values of **a** and **b**
and varying the value of **k**.

If we let **a** = 1 and **b** = 1, we get the equations

When **t** = 0, we get the equations

which define the point (1,1). We can see that for any value
of **k**, the graph will pass through the point (1,1) when
**t** = 0. Since **t** = 0 will be a common point for all
values of **k**, we will set the range for **t** so that
0 falls somewhere in the middle. We will try values between -1
and 1.

If we let **k** = 1, we get the following equations and
graph:

It appears to be a line with a slope of 1.

If we let **k** = 2 and graph the equation over the first
one, we get the following equations and graph:

The new graph appears to be a line with a slope of 2.

We see that the two graphs share the point (1,1) as expected.

If we let **k** = 3 and graph the equation over the first
two, we get the following equations and graph:

The new graph appears to be a line with a slope of 3.

We see, again, that the graph passes through the point (1,1) as expected.

Looking at our original equations

it seems that they represent a line with a slope of **k**
that passes through the point (**a**,**b**).

We will now try different values for **a**, **b**, and
**k** to see if this holds true.

If we let **a** = -2, **b** = 1, and **k** = -1, we
get the following equations and graph

We see that the graph has a slope of -1 and passes through the point (-2,1) as predicted.

If we let **k** = -3 and graph the equation over the first
one, we get the following equations and graph:

We see that the graph has a slope of -3 and passes through the point (-2,1) as predicted.

And finally, if we let **k** = 2 and graph the equation
over the first two, we get the following equations and graph:

We see that the graph has a slope of 2 and passes through the point (-2,1) as predicted.