Reciprocals of Quadratic Functions with One Zero or No Zeros

When the parabola only touches the axis, or misses it

In the last lesson the parabola crossed the x-axis twice. But a parabola can also touch the axis once (like y = (x − 2)²) or miss it completely (like y = x² + 1). Each case gives a very different reciprocal.

The discriminant b² − 4ac, which you used in the MCR3U unit Quadratic Functions, tells you which case you have before you draw anything:

• b² − 4ac > 0: two zeros, so two vertical asymptotes (last lesson);
• b² − 4ac = 0: one double zero, so one vertical asymptote;
• b² − 4ac < 0: no zeros, so no vertical asymptotes.

Case 1: one double zero

Take f(x) = (x − 2)². It is zero at x = 2, so g(x) = 1/(x − 2)² has a vertical asymptote x = 2.

But f(x) is never negative: it is positive on both sides of x = 2. So the reciprocal is positive on both sides too. As x → 2⁻ and as x → 2⁺, (x − 2)² is a tiny positive number, so g(x) → +∞ from both sides.

Compare this with y = 1/(x − 2), where the two sides go opposite ways. This is a useful rule:

• at a simple zero of the denominator, the graph goes to opposite infinities on the two sides;
• at a double zero, it goes to the same infinity on both sides.

Case 2: no zeros

Take f(x) = x² + 4x + 5. Its discriminant is 16 − 20 = −4 < 0, so it has no zeros. Complete the square: f(x) = (x + 2)² + 1. Its smallest value is 1, at x = −2.

Since f(x) ≥ 1 for every x:

• the reciprocal is defined everywhere: no vertical asymptotes, domain ℝ;
• 1/f(x) is positive and at most 1/1 = 1, with the maximum at x = −2: maximum point (−2, 1);
• as x → ±∞, f(x) grows without bound, so 1/f(x) → 0: horizontal asymptote y = 0;
• range (0, 1].

The graph is a smooth 'hill' or 'bell-shaped bump' that never touches the x-axis. If the parabola opens down and never reaches the x-axis (for example −(x² + 2)), everything flips: the reciprocal is a 'valley' below the x-axis with a minimum.