# 3.3 Quadratic polynomial function  (Page 2/3)

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${y}_{\mathrm{max}}=-\frac{D}{4a}\phantom{\rule{1em}{0ex}}\text{at}\phantom{\rule{1em}{0ex}}x=-\frac{b}{2a}$

${y}_{\mathrm{min}}⇒-\infty$

Clearly, range of the function is (-∞, -D/4a].

Problem : Determine range of $f\left(x\right)=-3{x}^{2}+2x-4$

Solution : The determinant of corresponding quadratic equation is :

$D={b}^{2}-4ac=4-4X\left(-3\right)X\left(-4\right)=4-48=-44\phantom{\rule{1em}{0ex}}⇒D<0$

$a=-3\phantom{\rule{1em}{0ex}}⇒a<0$

The graph of function is parabola opening down. Its vertex represents the maximum function value. The maximum and minimum values of function are given by :

$⇒{y}_{\mathrm{max}}=-\frac{D}{4a}=-\frac{-44}{4X-3}=-\frac{44}{12}=-\frac{11}{3}$

${y}_{\mathrm{min}}⇒-\infty$

Range = (-∞, -11/3)

The discriminant of corresponding quadratic equation and coefficient of term “ ${x}^{2}$ ” of quadratic function together determine nature of quadratic function and hence its graph. Graphs of quadratic function is intuitive and helpful to remember results. As a matter of fact, we can interpret all properties of quadratic function, if we can draw its graph.

## Case 1 : d<0

If D<0, then roots are complex conjugates. It means graph of function does not intersect x-axis. If a>0, then parabola opens up. The value of quadratic function is positive for all values of x i.e.

$\phantom{\rule{1em}{0ex}}D<0,a>0\phantom{\rule{1em}{0ex}}⇒f\left(x\right)>0\phantom{\rule{1em}{0ex}}\text{for}\phantom{\rule{1em}{0ex}}x\in R$

If a<0, then parabola opens down. The value of quadratic function is negative for all values of x i.e.

$\phantom{\rule{1em}{0ex}}D<0,a<0\phantom{\rule{1em}{0ex}}⇒f\left(x\right)<0\phantom{\rule{1em}{0ex}}\text{for}\phantom{\rule{1em}{0ex}}x\in R$

Sign rule : If D<0, then sign of function is same as that of “a” for all values of x in R.

## Case 2 : d=0

If D=0, then roots are equal and is given by –b/2a. It means graph of function just touches x-axis. If a>0, then parabola opens up. The value of quadratic function is non-negative for all values of x i.e.

$D=0,a>0\phantom{\rule{1em}{0ex}}⇒f\left(x\right)\ge 0\phantom{\rule{1em}{0ex}}\text{for}\phantom{\rule{1em}{0ex}}x\in R$

If a<0, then parabola opens down. The value of quadratic function is non-positive for all values of x i.e.

$D=0,a<0\phantom{\rule{1em}{0ex}}⇒f\left(x\right)\le 0\phantom{\rule{1em}{0ex}}\text{for}\phantom{\rule{1em}{0ex}}x\in R$

Sign rule : If D=0, then sign of function is same as that of “a” for all values of x in R except at x=-b/2a, at which f(x)=0. We do not associate sign with zero.

## Case 3 : d>0

If D>0, then roots are unequal and are given by (–b±D)/2a. It means graph of function intersects x-axis at α and β (β>α). If a>0, then parabola opens up. The value of quadratic function is positive for all values of x in the interval (-∞,α) U (β,∞).The values of quadratic function are zero for values of x ∈{α,β}. The value of quadratic function is negative for all values of x in the interval (α,β).

$D>0,a>0\phantom{\rule{1em}{0ex}}⇒f\left(x\right)>0\phantom{\rule{1em}{0ex}}\text{for}\phantom{\rule{1em}{0ex}}x\in \left(-\infty ,\alpha \right)\cup \left(\beta ,\infty \right)\phantom{\rule{1em}{0ex}}⇒\text{Sign of function same as that of “a”}$

$D>0,a>0\phantom{\rule{1em}{0ex}}⇒f\left(x\right)=0\phantom{\rule{1em}{0ex}}\text{for}\phantom{\rule{1em}{0ex}}x\in \left\{\alpha ,\beta \right\}\phantom{\rule{1em}{0ex}}$

$D>0,a>0\phantom{\rule{1em}{0ex}}⇒f\left(x\right)<0\phantom{\rule{1em}{0ex}}\text{for}\phantom{\rule{1em}{0ex}}x\in \left(\alpha ,\beta \right)\phantom{\rule{1em}{0ex}}⇒\text{Sign of function opposite to that of “a”}$

If a<0, then parabola opens down. The value of quadratic function is positive for all values of x in the interval (α,β).The values of quadratic function are zero for values of x ∈{α,β}. The value of quadratic function is negative for all values of x in the interval (-∞,α) U (β,∞).

$D>0,a<0\phantom{\rule{1em}{0ex}}⇒f\left(x\right)<0\phantom{\rule{1em}{0ex}}\text{for}\phantom{\rule{1em}{0ex}}x\in \left(\alpha ,\beta \right)\phantom{\rule{1em}{0ex}}⇒\text{Sign of function same as that of “a”}$

$D>0,a<0\phantom{\rule{1em}{0ex}}⇒f\left(x\right)=0\phantom{\rule{1em}{0ex}}\text{for}\phantom{\rule{1em}{0ex}}x\in \left\{\alpha ,\beta \right\}\phantom{\rule{1em}{0ex}}$

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