Quadratic expression concept on inequality - I

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Quadratic expression where the maximum degree of equation is 2

the most general form is,
y = a{x^2} + bx + c , a \ne 0 , a is not equal to zero because if a is zero it will not be a quadratic expression

or, f(x) = a{x^2} + bx + c,a \ne 0 ,  y = f(x) means y is a function of x

lets understand,
y = {x^2}

when x = 2, y = 4 and x = -2 , also y = 4
x = 3, y = 9 and x = -3 , y = 9
......................................

this means for two equal and opposite values of x , y has the same value
and y will always be positive because {x^2} is always positive. The graph will be concave upwards and no part of the curve will be below the x-axis as y is always positive. further, when x = 0 , y = 0 and for y = 4 which is represent by a straight line parallel to x-axis and at distance of 4 units from x-axis , cut the graph on two points i.e. x = 2 and -2

means curve  y = 4 and y = {x^2} have two common points -2 and 2



therefore for , y = {x^2} => y \ge 0 \forall x as {x^2} \ge 0 (\forall  means for every)

Now, consider y = - {x^2}

since {x^2} \ge 0 , therefore  - {x^2} \le 0 => y \le 0 \forall x

Please note if a \ge b =>  - a \le - b
means if we multiply both side with -1 inequality gets reversed
3 \ge 2 \Rightarrow - 3 \le - 2


therefore curve of y = - {x^2} is concave downwards or convex upwards and no part of curve is above the x-axis because y \le 0 means curve will be on the negative side of y-axis or under x-axis whatever be the value of x.

 so if y = a{x^2} , a \ne 0
for a > 0 , curve is concave upwards and for a < 0 curve is concave downwards

Now consider the general form,

y = a{x^2} + bx + c or f(x) = a{x^2} + bx + c , a \ne 0

 if a > 0 , curve will be concave upwards
and if a < 0 , curve will be concave downwards

it is a general quadratic expression, and curve represents parabola
f(x) = a{x^2} + bx + c, a \ne 0
= a\left( {{x^2} + \dfrac{{bx}}{a} + \dfrac{c}{a}} \right)
= a\left( {{x^2} + \dfrac{{bx}}{a} +\underbrace{ {{\left( {\dfrac{b}{{2a}}} \right)}^2}}_{\rm{Comment:}} - {{\left( {\dfrac{b}{{2a}}} \right)}^2} + \dfrac{c}{a}} \right)
Comment: This term is added and subtracted to form a perfect square
= a\left\{ {{{\left( {x + \dfrac{b}{{2a}}} \right)}^2} - \dfrac{{{b^2}}}{{4{a^2}}} + \dfrac{a}{c}} \right\}
= a{\left( {x + \dfrac{b}{{2a}}} \right)^2} - \dfrac{{\left( {{b^2} - 4ac} \right)}}{{4a}}
= a{\left( {x + \dfrac{b}{{2a}}} \right)^2} - \dfrac{D}{{4a}}\ldots(1)
where D = {b^2} - 4ac is called the discriminant of the quadratic expression.
The roots of f(x)
To find the roots we simply put f\left( x \right) = 0: (why???)
f\left( x \right) = a{\left( {x + \dfrac{b}{{2a}}} \right)^2} - \dfrac{D}{{4a}} = 0
\Rightarrow \,\,\,\, a{\left( {x + \dfrac{b}{{2a}}} \right)^2} = \dfrac{D}{{4a}}
\Rightarrow \,\,\,\, x + \dfrac{b}{{2a}} = \pm \dfrac{{\sqrt D }}{{2a}}
\Rightarrow \,\,\,\, x = \dfrac{{ - b \pm \sqrt D }}{{2a}}
Thus, we see that there are two roots of f(x)
{x_1} = \dfrac{{ - b - \sqrt D }}{{2a}},\,\,{x_2} = \dfrac{{ - b + \sqrt D }}{{2a}}
*Roots of any equation are those value of x for which f(x) = 0 or y = 0
and on x-axis the value of y = 0... in fact equation of x -axis is y = 0







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