☰ Year plan · Mathematics

The sign of a function and quadratic inequalities

Week 4 Β· The sign of a constant, linear and quadratic function, and quadratic inequalities in one unknown Β· 3 lessons
2

The sign of a quadratic function in the three cases bΒ² βˆ’ 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 Β· the values of x smaller than the smaller root or larger than the larger rootbetween the roots Β· the values of x in the interval from one root to the other
Explanation

The sign of f(x) = axΒ² + 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 bΒ² βˆ’ 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 bΒ² βˆ’ 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 bΒ² βˆ’ 4ac < 0 the parabola never meets the axis, so f has the sign of a for every real x. Example: f(x) = xΒ² βˆ’ 2x βˆ’ 8 = (x + 2)(x βˆ’ 4), roots βˆ’2 and 4, a = 1 > 0: positive on ]βˆ’βˆž, βˆ’2[ and on ]4, ∞[, negative on ]βˆ’2, 4[.

xyβˆ’4βˆ’3βˆ’2βˆ’1123456βˆ’10βˆ’8βˆ’6βˆ’4βˆ’224680f(x) = xΒ² βˆ’ 2x βˆ’ 8βˆ’24
Worked examples
f(x) = xΒ² βˆ’ 2x βˆ’ 8. bΒ² βˆ’ 4ac = 4 + 32 = 36 > 0, roots βˆ’2 and 4. Positive on ]βˆ’βˆž, βˆ’2[ βˆͺ ]4, ∞[ (also written R βˆ’ [βˆ’2, 4]), negative on ]βˆ’2, 4[, zero at βˆ’2 and 4. Check: f(0) = βˆ’8 < 0 lies between the roots βœ“.
f(x) = βˆ’xΒ² + 6x βˆ’ 5 = βˆ’(x βˆ’ 1)(x βˆ’ 5). Roots 1 and 5, and a = βˆ’1 < 0, so the parabola opens down: f is positive between the roots on ]1, 5[ and negative on ]βˆ’βˆž, 1[ βˆͺ ]5, ∞[.
xyβˆ’112345βˆ’112340f(x) = xΒ² βˆ’ 6x + 13
f(x) = xΒ² βˆ’ 6x + 13. bΒ² βˆ’ 4ac = 36 βˆ’ 52 = βˆ’16 < 0 and a = 1 > 0, so f(x) > 0 for every real x. (The vertex is (3, 4), above the axis.)
f(x) = βˆ’9xΒ² + 12x βˆ’ 4 = βˆ’(3x βˆ’ 2)Β². The discriminant is 144 βˆ’ 144 = 0, root x = 2/3, a < 0. So f(x) < 0 for every x β‰  2/3 and f(2/3) = 0.
Class activity

Sketch it. On squared paper sketch the parabola of f(x) = xΒ² βˆ’ 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) = xΒ² βˆ’ 5x + 6.
2
Find the sign of f(x) = xΒ² + 4x + 5.
3
Find the sign of f(x) = 4xΒ² βˆ’ 4x + 1.
4
Find the sign of f(x) = 4 βˆ’ xΒ².
5
Find the sign of f(x) = βˆ’2xΒ² + x βˆ’ 3.