Functions Set 2
Example – 3
Find the roots of the function \(f(x) = a{x^2} + bx + c\) and plot its graph. Also solve the inequalities \(f(x) \ge 0\) and \(f(x) \le 0\)
Solution: The expression for \(f(x)\) is quadratic, of which we have seen some examples. Lets discuss a particular case for \(f(x)\) and then go on to the general case above.
Let \(f(x) = 3{x^2} + 10x + 3\) which can be written as \(f(x) = (3x + 1)(x + 3).\) This is 0 when either \(3x + 1 = 0\) or \(x + 3 = 0\)
\[i.e.\qquad \qquad {\rm{ }}x = - \frac{1}{3},{\mkern 1mu} {\mkern 1mu} - 3{\rm{ }}\qquad \qquad\qquad ...(i)\]
Now \(f(x) = 3{x^2} + 10x + 3\) can be rearranged as
We can obtain the graph for f(x) in the following steps:
We see that the coordinates of the vertex (the point v) are \(\begin{align}\left( { - \frac{5}{3}, - \frac{{16}}{3}} \right)\end{align}\) , and the graph crosses the axis at (it has its roots at) \(\begin{align}x = - 3, - \frac{1}{3}\end{align}\) as we have determined in (i) above. The graph goes below the axis in the interval \(\begin{align}\left( { - 3, - \frac{1}{3}} \right)\end{align}\) so that in this interval f(x) < 0. In the intervals \(\left( {-\infty ,{\rm{ }}-3} \right)\) and \(\left( { - \frac{1}{3},\infty } \right)\) , f(x) > 0. Now we know everything about this quadratic function
Hence, f (x) < 0 between the roots and f(x) > 0 outside this interval.
Now we apply this reasoning to
\[\begin{align}&f(x) = a{x^2} + bx + c\\&f(x) = a{x^2} + bx + c = a\left( {{x^2} + \frac{{bx}}{a} + \frac{c}{a}} \right)\end{align}\]
\[\begin{align}&= a\left( {{x^2} + \frac{b}{a}x + {{\left( {\frac{b}{{2a}}} \right)}^2} - {{\left( {\frac{b}{{2a}}} \right)}^2} + \frac{c}{a}} \right)\\&= a\left( {{{\left( {x + \frac{b}{{2a}}} \right)}^2} - \frac{{{b^2} - 4ac}}{{4{a^2}}}} \right)\\&= a{\left( {x + \frac{b}{{2a}}} \right)^2} - \frac{D}{{4a}} \end{align}\]
where \(D = {b^2} - 4ac\) is called the discriminant of the quadratic expression.
\[f(x) = 0\;{\rm{ when \;}}a\;{\left( {x + \frac{b}{{2a}}} \right)^2} = \frac{D}{{4a}}\]
\[ \Rightarrow x + \frac{b}{{2a}} = {\rm{ }}\frac{{ \pm \sqrt D }}{{2a}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,x = - \frac{{b \pm \sqrt D }}{{2a}}\]
\(\begin{align}{\rm{The \;two\; roots \;of\; }}f(x){\rm{\; }}are{\rm{\; }}{x_1} = \frac{{ - b - \sqrt D }}{{2a}},{x_2} = \frac{{ - b + \sqrt D }}{{2a}}\;\;\;\;\;\;\;\;\qquad\qquad {\rm{}}...(ii)\end{align}\)
The graph is obtained in the following steps:
\[{x^2}\rightarrow{{}}a{x^2}\rightarrow{{}}a{\left( {x + \frac{b}{{2a}}} \right)^2}\rightarrow{{}}a{\left( {x + \frac{b}{{2a}}} \right)^2} - \frac{D}{{4a}}\]
We see that if a < 0, the graph will not open upwards but downwards
Also, whether a > 0 or a < 0, the co-ordinates of the vertex are \(\begin{align}\left( { - \frac{b}{{2a}},\frac{{ - D}}{{4a}}} \right)\end{align}\) . What if a > 0 and D < 0? The y-co-ordinate of the vertex, \(\begin{align}\frac{{ - D}}{{4a}}\end{align}\) is positive, and hence the vertex lies above the x-axis.
f (x) in this case never cuts the axis and hence never becomes 0. We say that f(x) has no real roots. f(x) > 0 for all values of x.
Similarly, if a < 0 and D < 0, the y-co-ordinate of the vertex is negative and the graph lies below the axis.
Here also, f(x) has no real roots.
We see that whether a > 0 or a < 0, if D < 0, f(x) will not have real roots.
This is also evident in the formula for the roots in (ii). If D < 0, then \(\sqrt D \) is imaginary (non-real).
What if a > 0 and D = 0? In this case, the y-coordinate of the vertex is 0 or the vertex lies on the x-axis.
Now we can easily determine the solutions to f(x) > 0 and f (x) < 0
All these results are summarized below.
You are urged not to memorize the results but understand them, by verifying each of them on your own