3.3 Quadratic polynomial function (Page 3/3)

Functions

Page 3 / 3

$D > 0, a < 0 \Rightarrow f (x) > 0 for x \in (- \infty, α) \cup (β, \infty) \Rightarrow Sign of function opposite to that of “a”$

Sign rule : If D>0, then domain of function, which is R, is divided at root points in three intervals. The signs of function in side intervals are same as that of “a”, whereas sign of function in the middle interval is opposite to that of “a”.

Examples

Problem : Determine interval of “a” for which graph of $x^{2} + (a - 1) x + 16$ lie above x-axis.

Solution : Here coefficient of “ $x^{2}$ ” is positive. Now, the graph of quadratic function lie above x-axis when D>0 and a>0.

$D = {(a - 1)}^{2} - 4 X 1 X 16 < 0 \Rightarrow {(a - 1)}^{2} - 64 < 0$

$\Rightarrow {(a - 1)}^{2} - 8^{2} < 0 \Rightarrow (a - 1 + 8) (a - 1 - 8) < 0$

$\Rightarrow (a + 7) (a - 9) < 0$

$\Rightarrow - 7 < a < 9$

Problem : The graph of a quadratic expression $f (x) = a x^{2} + b x + c$ is shown in the figure. Determine signs of a,b,c.

Solution : The parabola opens downward. It means a<0. Since both roots are positive, their sum is also positive.

$α + β = - \frac{b}{a} > 0 \Rightarrow Signs of a and b are opposite.$

It means b>0. Now, putting x=0 in the function, we have value of function as :

$f (0) = a X 0^{2} + b X 0 + c = c$ From figure, graph can intersects y-axis only at negative y-value. Hence, “c” is negative.

Quadratic inequality

Value of quadratic function is positive, zero or negative depending on the nature of coefficient of “ $x^{2}$ ” and discriminant, D. Accordingly, a quadratic function will hold true for particular interval(s) depending upon these parameters. Consider an example :

$\Rightarrow 2 x^{2} - 5 x - 3 > 0$

$\Rightarrow 2 x^{2} - 6 x + x - 3 > 0$

$\Rightarrow 2 x (x - 3) + (x - 3) > 0$

$\Rightarrow (2 x + 1) (x - 3) > 0$

$\Rightarrow x = - \frac{1}{2}, 3$

Here,

$a = 2 \Rightarrow a > 0$

$D = 25 - 4 X 2 X 3 = 1 \Rightarrow D > 0$

Under these conditions, given quadratic function is positive for $(- \infty, - 1 / 2) \cup (3, \infty)$ . On the real number line, the intervals are shown as :

Let us consider slightly changed inequality involving less than equal sign,

$2 x^{2} - 5 x - 3 \leq 0$

Again quadratic function is positive for $(- \infty, - 1 / 2) \cup (3, \infty)$ . However, equality is also allowed. It means valid intervals should also include root points. The modified valid interval corresponding to "less than equal to inequality" is : $(- \infty, - 1 / 2] \cup [3, \infty)$ .

From this illustration, it is clear that we can determine valid interval(s) of x, provided we know the signs of quadratic function in different intervals. If equality is also allowed as in the case of “less than equal to” or “greater then equal to”, then we need to include root points also.

Problem : Determine the interval of x for which $f (x) = 2 x^{2} - 5 x - 3$ is non-positive and negative.

Solution : As already determined earlier, roots of function are -1/2, 3. Also a>0 and D>0. Sign rule for the function is shown here :

From the figure, it is clear that function is non-positive i.e. f(x) ≤0 in the interval [-1/2, 3]. Also, function is negative i.e. f(x)<0 in the interval (-1/2, 3).

Problem : A quadratic function is given by $f (x) = x^{2} - 4 x + 4$ . Find solution for each of four inequalities viz f(x)<0, f(x) ≤ 0, f(x)>0 and f(x) ≥ 0.

Solution : Here, coefficient of “ $x^{2}$ ” is 1. Thus,

$\Rightarrow a > 0$

Determinant of corresponding quadratic equation is :

$D = {(- 4)}^{2} - 4 X 1 X 4 = 0$

For D=0 and a>0, f(x) ≥0. Further, since D=0, it means that corresponding quadratic equation has one real root. Now, the root is :

$α = - \frac{b}{2 a} = - \frac{- 4}{2 X 1} = 2$

This means that f(x) is positive for all values of x except for x=2. At x=2, f(x)=0. It follows then that :

$f (x) \geq 0; x \in R$ $f (x) > 0; x \in R - {2}$ $f (x) \leq 0; x \in {2}$ $f (x) < 0; No solution$

A quadratic function is given by $f (x) = - x^{2} + 2 x - 4$ . Find solution for each of four inequalities viz f(x)<0, f(x) ≤ 0, f(x)>0 and f(x) ≥ 0.

Here, coefficient of “ $x^{2}$ ” is -1. Thus,

$\Rightarrow a < 0$

Determinant of corresponding quadratic equation is :

$D = {(2)}^{2} - 4 X - 1 X - 4 = - 8 \Rightarrow D < 0$

For D<0 and a<0, f(x)<0. Further, since D<0, it means that corresponding quadratic equation has no real root. This means that f(x) is negative for all values of x. It follows then that

$f (x) < 0; x \in R$ $f (x) \leq 0; x \in R$ $f (x) > 0; No solution$ $f (x) \geq 0; No solution$

Acknowledgment

Author wishes to thank Mr. Ralph David for helping to correct a mistake in the module.

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Source: OpenStax, Functions. OpenStax CNX. Sep 23, 2008 Download for free at http://cnx.org/content/col10464/1.64

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3.3 Quadratic polynomial function (Page 3/3)

Sign rule of quadratic function

Examples

Graph of quadratic function

Quadratic inequality

Quadratic inequality

Quadratic inequality

Sign rule of quadratic inequality

Acknowledgment

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