Question

Explain the method of finding maximum and minimum of quadratic functions.

Answer 1

Hint : We are supposed to find the method of finding maximum and minimum values of quadratic functions having a form of $a{x^2} + bx + c$ so we have three methods of doing so. We take the first method with an example and apply the method to see how we can get the maxima and minima of the quadratic functions.

Complete step-by-step answer :
We know that a quadratic function is in the form
$a{x^2} + bx + c = f(x)$
We start with what is called as beginning with the general form of function where we write the function and see if it is in the general standard form. We take an example of quadratic function
$2{x^2} + 5x + 4$
We know that a quadratic function when drawn on a graph gives us a parabola which either opens upward or downward. So for determining the direction of the graph of this parabola we see the coefficient of ${x^2}$ term. If it is positive then the graph is upward and if it is negative it is downward. In our case the coefficient is +2 which results in an upward parabola.
When the graph is upward we will check for its minimum value and if it is downward we check for its maximum value like shown in the below figure -

Now we compare our function with the standard quadratic function and have
$2{x^2} + 5x + 4 = a{x^2} + bx + c \\\ \Rightarrow a = 1,b = 5,c = 4 \;$
In the next step we calculate $- \dfrac{b}{{2a}}$ which tells us the value of $x$ vertex of the parabola i.e.
$x = - \dfrac{b}{{2a}}$
Using this we find the minimum value of our parabola
$x = - \dfrac{5}{4}$
We have got one vertex of the graph not to get the other one we put the value of $x$ into our function to get the value of $y$ i.e.
$f(x) = 2{x^2} + 5x + 4 \\\ \Rightarrow f\left( { - \dfrac{5}{4}} \right) = 2{\left( { - \dfrac{5}{4}} \right)^2} + 5\left( { - \dfrac{5}{4}} \right) + 4 = {2} \times \dfrac{{25}}{{{{\mathop {16}\limits_8 }}}} - \dfrac{{5 \times 5}}{4} + 4 = \dfrac{{25}}{8} - \dfrac{{25}}{4} + \dfrac{4}{1} \\\ \Rightarrow f\left( { - \dfrac{5}{4}} \right) = \dfrac{{25 - 50 + 32}}{8} = \dfrac{7}{8} \;$
So our minimum value of the function will be at the following point:
${\text{Minimum}}\,{\text{value}} = \left( { - \dfrac{5}{4},\dfrac{7}{8}} \right)$

Note : To calculate the maximum value of a quadratic function we use the same method where we take $- \dfrac{b}{{2a}}$ and see what our $x$ comes out, then in the same manner find the value of $y$ to get the point of maxima.