3.3 Quadratic polynomial function  (Page 3/3)

Problem : Determine interval of “a” for which graph of $x^2+\left(a-1\right)x+16$ 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.

$$D={\left(a-1\right)}^2-4X1X16<0\Rightarrow {\left(a-1\right)}^2-64<0$$

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

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

$$\Rightarrow -7<a<9$$

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

Graph of quadratic function

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

$$\alpha +\beta =-\frac{b}{a}>0\Rightarrow \text{Signs of a and b are opposite.}$$

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

$$f\left(0\right)=aX{0}^2+bX0+c=c$$ From figure, graph can intersects y-axis only at negative y-value. Hence, “c” is negative.

Problem : Determine the interval of x for which $f\left(x\right)=2x^2-5x-3$ 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 :

Sign rule of quadratic inequality

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 $f\left(x\right)=x^2-4x+4$ . 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 :

$$D={\left(-4\right)}^2-4X1X4=0$$

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 :

$$\alpha =-\frac{b}{2a}=-\frac{-4}{2X1}=2$$

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 :

$$f\left(x\right)\ge 0;x\in R$$ $$f\left(x\right)>0;x\in R-\left\{2\right\}$$ $$f\left(x\right)\le 0;x\in \left\{2\right\}$$ $$f\left(x\right)<0;\text{No solution}$$