# show that 2x-14 is the tangent of x^2-6x +2 without knowledge of calculus?

Relevance
• If the line y = 2x - 14 intersects the parabola y = x^2 - 6x + 2 then

x^2 - 6x + 2 = 2x – 14, or

x^2 - 8x + 16 = 0 = (x – 4)^2

But that indicates that there is only one, (coincident) point of intersection.

That perfect square is the characteristic of the line being a tangent.

• Show that line L: y = 2x-14...(1) is tangent to parabola P: y = x^2-6x+2...(2).;

Solve (1) & (2) for the intersection of L & P. Substitute (1) in (2): 2x-14 = x^2-6x+2, ie., x^2-8x+16 = 0, ie., (x-4)^2 = 0, ie., x = 4 {multiplicity 2}. Therefore L intersects P

in only 1 place. L is tangential to P at (4,-6).

• Neither 2x - 14 nor x² - 6x + 2 defines a curve. For that you would need to have some sort of relations in at least two variables, perhaps equations. Simple expressions in one variable will not do. Having no curves, you can have no tangency.

• If the line is tangent to the curve, the first degree line and second degree curve will intersect in one and only one place.

2x - 14 = x² - 6x + 2

x² - 8x + 16 = 0

If the line and parabola are tangent, the discriminant of the quadratic will be zero (only one unique solution).

b² - 4ac = 0

64 - 4*16 = 0

0 = 0

Further proof will be to show there is only 1 unique solution of the quadratic.

x² - 8x + 16 = 0

(x-4)² = 0

x = 4

QED