The Product of Two Linear Functions
by Emily Kennedy

If we multiply two linear functions,
f(x) = mx + j and g(x) = nx + k (with m, n ≠ 0),
we get a quadratic function h(x).

Play with the values of m, j, n, and k in this Graphing Calculator file. Make some observations about the points of intersection of
f(x), g(x), and h(x).
Can you find values that give you something like this?

Let's find two linear functions f(x) and g(x) such that
f(x) and g(x) are tangent to h(x) at
(a, ya) and (b, yb), where a ≠ b.

For f(x) to be tangent to h(x) at (a, ya),
we must have
ya = f(a) = h(a) and f '(a) = h '(a).

And for g(x) to be tangent to h(x) at (b, yb),
we must have
yb = g(b) = h(b) and g '(b) = h '(b).

ya = f(a) = h(a) = f(a)g(a) f(a) = 0 or g(a) = 1 (*)

yb = g(b) = h(b) = f(b)g(b) g(b) = 0 or f(b) = 1 (**)

We now have four (not necessarily mutually exclusive) possibilities:

 Case (from *) (from **) f(a) g(a) f(b) g(b) A 0 1 B 1 0 C 0 0 D 1 1

Let's deal with each case separately and see what happens.

Case A: f(a) = 0 and f(b) = 1

(since m ≠ 0 by assumption)

(since m ≠ 0 by assumption)

Let's fill in our chart:

 Case (from *) (from **) f(a) g(a) f(b) g(b) A 0 1 1 0 B 1 0 C 0 0 D 1 1

So Case A Cases B, C, and D.
Are all four cases truly equivalent, or are there some functions f and g
which satisfy Case B, C, or D but not A?
Let's find out!

Case B: g(a) = 1 and g(b) = 0

(since n ≠ 0 by assumption)

(since n ≠ 0 by assumption)

Let's fill in our chart:

 Case (from *) (from **) f(a) g(a) f(b) g(b) A 0 1 1 0 B 0 1 1 0 C 0 0 D 1 1

So Cases A and B are equivalent. What about the other two cases?

Case C: f(a) = 0 and g(a) = 1

(since m ≠ 0 by assumption)

(since n ≠ 0 by assumption)

Let's fill in our chart:

 Case (from *) (from **) f(a) g(a) f(b) g(b) A 0 1 1 0 B 0 1 1 0 C 0 1 1 0 D 1 1

So Cases A, B, and C are all equivalent.

Finally, let's check out Case D.

Case D: g(a) = 1 and f(b) = 1

(since n ≠ 0 by assumption)

(since m ≠ 0 by assumption)

Let's fill in our chart:

 Case (from *) (from **) f(a) g(a) f(b) g(b) A 0 1 1 0 B 0 1 1 0 C 0 1 1 0 D 0 1 1 0

So all four cases (A, B, C, and D) are equivalent!

That is, anytime we have functions
f(x), g(x), and h(x) = f(x)g(x) such that
f(x) is tangent to h(x) at (a,ya)
and g(x) is tangent to h(x) at (b,yb),
we have:
f(a) = 0, g(a) = 1, f(b) = 1, and g(b) = 0.

Click here to find the values of m, j, and n, and k
that give us functions that satistfy these properties.