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The sign of a function and quadratic inequalities

الأسبوع 4 · إشارة الدالة الثابتة والخطية والتربيعية، ومتباينات الدرجة الثانية في مجهول واحد · 3 دروس
2

The sign of a quadratic function in the three cases b² − 4ac > 0, = 0 and < 0, read from the parabola

Mathematics Sec 1 Term 1, Unit 1, Lessons 4–5
GoalFind the sign of a quadratic function from its roots and the sign of a.
New words
outside the roots · the values of x smaller than the smaller root or larger than the larger rootbetween the roots · the values of x in the interval from one root to the other
Explanation

The sign of f(x) = ax² + bx + c follows the parabola: above the x-axis f is positive, below it f is negative, on it f is zero. The discriminant tells how the parabola meets the axis. If b² − 4ac > 0 there are two roots L < M. The parabola opens up when a > 0, so it is above the axis outside the roots and below it between them: f has the sign of a outside [L, M] and the opposite sign between L and M. If b² − 4ac = 0 the parabola only touches the axis at one point, so f has the sign of a everywhere except at that root, where it is zero. If b² − 4ac < 0 the parabola never meets the axis, so f has the sign of a for every real x. Example: f(x) = x² − 2x − 8 = (x + 2)(x − 4), roots −2 and 4, a = 1 > 0: positive on ]−∞, −2[ and on ]4, ∞[, negative on ]−2, 4[.

xy−4−3−2−1123456−10−8−6−4−224680f(x) = x² − 2x − 8−24
Worked examples
f(x) = x² − 2x − 8. b² − 4ac = 4 + 32 = 36 > 0, roots −2 and 4. Positive on ]−∞, −2[ ∪ ]4, ∞[ (also written R − [−2, 4]), negative on ]−2, 4[, zero at −2 and 4. Check: f(0) = −8 < 0 lies between the roots ✓.
f(x) = −x² + 6x − 5 = −(x − 1)(x − 5). Roots 1 and 5, and a = −1 < 0, so the parabola opens down: f is positive between the roots on ]1, 5[ and negative on ]−∞, 1[ ∪ ]5, ∞[.
xy−112345−112340f(x) = x² − 6x + 13
f(x) = x² − 6x + 13. b² − 4ac = 36 − 52 = −16 < 0 and a = 1 > 0, so f(x) > 0 for every real x. (The vertex is (3, 4), above the axis.)
f(x) = −9x² + 12x − 4 = −(3x − 2)². The discriminant is 144 − 144 = 0, root x = 2/3, a < 0. So f(x) < 0 for every x ≠ 2/3 and f(2/3) = 0.
Class activity

Sketch it. On squared paper sketch the parabola of f(x) = x² − 5x + 6 using only its roots and the sign of a, then write the sign of f in each part of the x-axis.

Practice
1
Find the sign of f(x) = x² − 5x + 6.
2
Find the sign of f(x) = x² + 4x + 5.
3
Find the sign of f(x) = 4x² − 4x + 1.
4
Find the sign of f(x) = 4 − x².
5
Find the sign of f(x) = −2x² + x − 3.