How do you find the points of intersection or tangency of a straight line with a parabola?

Answer

Intersections are the points whose coordinates satisfy both the equation of the parabola and the line. So set the two y values ​​equal and you get a quadratic equation in x that you can solve. That gives 3 possibilities:
1) two different real roots (=> two different points of intersection)
2) two equal roots (discriminant is zero) => the two points of intersection coincide => point of tangency
3) two complex roots (discriminant negated) => no real intersections

e.g. with the parabola and line 1:
-2x^2 + 2x = -2x
or even simplified:
x^2 – 2x + 0 = 0
so that: discriminant Δ = b^2 – 4ac = (-2)^2 – 4.1.0 = 4
so two points of intersection: x= 0 (for which y = 0) and x = 2 (with y = -4)

the sum of the x-coordinates of a square equation ax^2 + Bx + c = 0 is always -b/a

you can easily see that if you look at the way you calculate it, with the discriminant Δ :

first square root: x1= [ – b + wortel(Δ) ] / 2a
second square root: x2= [ – b – wortel(Δ) ] / 2a
so if you add x1 + x2 = -b/a

try it with the example above : x1 + x2 = 2 and indeed -b/a = -(-2)/1 = also 2

For the 2nd line you find two coincident intersection points, so a tangent point in x = 1/3

You can also check that it is a tangent point, because the derivative at that point is 2/3 and that is indeed the slope of line 2.

So at that point the x-coordinate is 1/3, although one could also argue that it is 2/3 since it is a double root, so it must be taken twice. That is also what you find when you use the formula “sum roots = -b/a”.