Assignment 1

The goal of this assignment is to find two distinct linear functions f(x) and g(x) so that they are tangent to their product h(x)=f(x)g(x) in exactly two places (like the picture below).

 

 

I started off using a method of trial and error by graphing two linear functions and then trying to make observations that would lead to a general case.

 

Trial 1:

For the first try, I selected two equations (at random) f(x)=x+2 and g(x)=x+4 making the product h(x)=(x+4)(x+2) and the graphs looks like this:

What can we note from our first try?

Trial 2:

Let's use similar values and generate a couple sets of equations. On the right f(x) = x+2, g(x) = -2x+4, and h(x) = (x+2)(-2x+4). On the left f(x) = x+2 and g(x) = -x+4 and h(x) = (x+2)(-x+4):

What can we note from our second try?

Trial 3:

We can get the parabola to be symmetric about the y-axis by making the roots of it eqaul and opposite. So let's graph the equations f(x) = x+2, g(x) = -x+2, and h(x) = (x+2)(-x+2)

What can we note from the third try?

x+2 = -x+2

2x+2 = 2

2x = 0

x = 0

Now let's find the x coordinate of the vertex of the parabola (the formula is x = -b/2a where h(x) = ax^2+bx+c)

h(x) = (x+2)(-x+2)

h(x) = -x^2+4

Vertex = -0/-2 = 0

So the x-coordinates of the intersection point and the vertex are equal. This in fact turns out to be true in the general case.

To see this click here for Proof 1.

 

Trial 4:

Now I will try to modify the above values of the y intercepts of the linear equations to see if I can get them to be tangent to the parabola (for details on the numerical values I choose look at proof 2). Let's try f(x) = x+3/4, g(x) = -x+3/4, and h(x) = (x+3/4)(-x+3/4).

 

What can we note from this try?

Trial 5: Let's try again with the equations f(x) = x+1/2, g(x) = -x+1/2, and h(x) = (x+1/2)(-x+1/2).

What can we note from trial 5?

x+1/2 = (x+1/2)(-x+1/2)

(x+1/2)/(x+1/2) = ((x+1/2)(-x+1/2))/(x+1/2)

1 = -x+1/2

x = -1/2

Great they intersect at -1/2 and 1/2 (you can get 1/2 by letting g(x) = h(x)). These are the values of the x-intercepts of the linear equations and the two roots of the quadratic.

Trial 6: Let's change the slope of the linear equations and see if our functions still work. Let f(x) = 2x+1/2, g(x) = -2x+1/2, and h(x) = (2x+1/2)(-2x+1/2).

What can we note from trial 6?

Trial 7:

Now let's check and see what happens when we change the y-intercpets of the linear equations. Let f(x) = x+3, g(x) = -x-2, and h(x) = (x+3)(-x-2) (left hand graph) and f(x) = x+3, g(x) = -x-1, and h(x) = (x+3)(-x-1) (right hand graph).

 

 

What can we note from trial 7?

 

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