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Just Touching

Part of the Quadratics & Complex Numbers section of Coddy's Algebra 2 journey. Lesson 35 of 48.

Sometimes the quadratic you get has only one solution. Take y=x2y = {x}^{2} and y=2x−1y = 2x-1. Substituting gives x2=2x−1{x}^{2} = 2x-1, so x2−2x+1=0{x}^{2}-2x+1 = 0, which factors as (x−1)2=0{\left(x-1\right)}^{2} = 0. The only x is 1, and the line gives y = 1: one meeting point, (1, 1).

A line like this is a tangent: it touches the parabola at one point without crossing it. Everywhere else the parabola stays above the line, because the difference between them is the square of x - 1, which is never negative.

The discriminant sees this coming. x2−2x+1=0{x}^{2}-2x+1 = 0 has discriminant 4−44-4, which is 0: one repeated solution, so one meeting point. A positive discriminant would have meant two crossings, as in the last lesson.

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Challenge

Find where the line y=4xy = 4x meets the parabola y=x2+4y = {x}^{2}+4.

Try it yourself

y=x2+4y = x^{2}+4y=4xy = 4x

Find xx, yy

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