Proof 1: The x coordinate of the vertex is the same as the x-coordinate of the point of intersection of the lines.

Assume from observation and rules of symmetry that the slopes of the linear equations must be equal and opposite.

Therefore f(x) = ax +b and g(x) = -ax+c for the point of intersection we must set them equal to one another.

f(x)=g(x)

ax+b = -ax+c

2ax = c-b

x = (c-b)/2a

So this is the x coordinate for the point of intersection. The x coordinate for the vertex is a little more sticky.

h(x) = (ax+b)(-ax+c)

h(x) = (-a^2)(x^2)+a(c-b)x+bc

then we get x = -a(c-b)/2(-a^2)

with simplification x = (c-b)/2a

And we are done. The x-coordinates must be the same if we assume the slopes are equal and opposite.

Proof 2: Show for what values of y the vertex of the parabola will be below the intersection point of the two linear equations.

Assume from earlier observations that the parabola has two real roots and therefore the y coordinate of the the parabola must be

greater than zero.

Find the y-coodinate of the intersection point of the lines. From above we have that x = (c-b)/2a so let's find the y coordinate

f((c-b)/2a) = a(c-b)/2a + b

f((c-b)/2a) = (c-b)/2 +b

f((c-b)/2a) = (c+b)/2

Let's find the y coordinate of the vertex of the parabola.

h((c-b)/2a) = (-a^2)((c-b)/2a)^2 + a(c-b)((c-b)/2a) + bc

h((c-b)/2a) = -1/4(c-b)^2 + 1/2(c-b)^2 + bc

h((c-b)/2a) = 1/4(c+b)^2 + bc

h((c-b)/2a) = ((c+b)/2)^2

So we want the y coordinate of the vertex to be less than the y coordinate of the intersection of the linear equations...((c+b)/2)^2 < (c+b)/2 or by simplification 0 < (c+b) < 2. It is interesting to note that the slope of the lines does not determine whether or not the vertex will be below the point of intersection of the lines. From this we can conclude that the only condition on the slopes of the lines is that they are equal and opposite. This along with proof 1 gives us the general coordinates of the vertex of the parabola and intersection point of the line.

Intersection Point: ((c-b)/2a, (c+b)/2)

Vertex Point: ((c-b)/2a, ((c+b)/2)^2)

Proof 3: We want to know the points of intersection of the graphs when h(x) = f(x)g(x). To do this let's set f(x) = h(x) or f(x) = f(x)g(x). The parabola and the linear graph intersect when

f(x) = f(x)g(x)

f(x)/f(x) = f(x)g(x)/f(x)

g(x) = 1

or when f(x) = 0 (as noted in trial 1)

If f(x) = 0 and g(x) = 1 at different x values then this leads to two intersection points per linear equation. So to get only one intersection point from the graphs we need f(x) = 0 and g(x) = 1 at the same x value. So we have that

f(x) + g(x) = 1

or ax+b+-ax+c = 1

b+c = 1

or c = 1-b

(The converse holds true as well or when g(x) = g(x)f(x) then g(x) = 0 and f(x) =1 and we get the same equation b + c = 1)

This now gives us precise coordinates of the y coordinate of the vertex as well as the y coordinate of the linear intersection.

Vertex Point: ((c-b)/2a, 1/2)

Intersection Point: ((c-b)/2a, 1/4)

This implies that our parabola and linear equations can slide up and down the x-axis depending on our choice of b and c but that the y point of the vertex and linear intersection must be fixed.

Necessary and Sufficient Conditions:

These are the two conditions for us to get tangency

I) The slopes of the linear equations are equal and opposite.

II) When you add the y intercepts of the linear equations they equal 1.