Maximum and Minimum Value of Quadratic Equation

Quadratic Equation IIT JEE

To find the minimum value of Quadratic Equation we first need to understand the nature of graph of the quadratic equation for different values of ‘a’. The graph of Quadratic equation f(x) = \(ax^{2} + bx + c\) will be either concave upwards or concave downwards i.e. if a > 0 or a < 0 respectively. When the graph of a quadratic equation is concave upwards then its vertex determines the minimum value of the quadratic function f(x) and when the graph of a quadratic equation is concave downwards, then its vertex determines the maximum values of the quadratic function f(x). These Quadratic Equation IIT JEE notes are not only helpful in solving several complicated numericals asked in IIT JEE but these study notes are even helpful in the preparation of various other engineering entrance examinations like VITEEE, BITSAT, SRMEEE, etc, quite easily.

  1. Introduction to the Quadratic Equation
  2. The Theory of Equations
  3. Graphing Quadratic Equations
  4. Intervals in which the roots of the quadratic equation lie
  5. Quadratic Inequalities

Maximum and Minimum Value of Quadratic Equation:

1: Absolute Range of a Quadratic Equation:

Case 1: When a > 0, the absolute range of a Quadratic Expression is given by:

\(\mathbf{\left [ -\;\frac{b^{2}\;-\;4ac}{4a},\;\infty \right )}\) or \(\mathbf{\left [ -\;\frac{D}{4a},\;\infty \right )}\)

Case 2: When a < 0, the absolute range of a Quadratic Equation is given by:

\(\mathbf{\left ( -\;\infty , \;-\;\frac{D}{4a}\right ]}\)

Also, the maximum and minimum values of quadratic equation f (x) occurs at x = \(\mathbf{-\;\frac{b}{2a}}\).

If the given Quadratic equation is in the form f (x) =\(a (x – h)^{2} + m\), Then the value of ‘m’ (vertex) gives us the minimum (when ‘a’ is negative) or maximum (when ‘a’ is positive) values of the given function.

Example: Find the maximum or minimum value of quadratic equation \(– 4 (x – 2)^{2} + 2\).

Solution:

From the above Quadratic Expression, a = – 4 and m = 2.

Since the value of ‘a’ is negative, therefore the given quadratic equation will have a maximum value.

Hence, the maximum value of quadratic equation \(– 4 (x – 2)^{2} + 2\) is 2.

Maximum and minimum values of Quadratic Equation: Range in Restricted Domain, when x \(\mathbf{\in }\) [m,n]:

Case 1: \(\mathbf{-\frac{b}{2a}\;\notin \;\left [ m\;,\;n \right ]}\)

f (x) = \(\mathbf{\in }\) [ minimum { f (m), f (n) }, maximum { f (m), f (n) } ]

Case 2: \(\mathbf{-\frac{b}{2a}\;\in \;\left [ m\;,\;n \right ]}\)

f (x) = \(\mathbf{\in \left [ minimum \;\left \{ f\left ( m \right ),\;f\left ( n \right ) \;-\;\frac{D}{4a}\right \}, \;maximum \;\left \{ f\left ( m \right ),f\left ( n \right ), -\frac{D}{4a} \right \}\right ]}\)

Example 1: Find the minimum and maximum values of quadratic equation f (x) = \(x^{2} – 12x + 11\).

Solution:

Since a > 0, the maximum and minimum values of a Quadratic Expression is given by:

\(\mathbf{\left [ -\;\frac{b^{2}\;-\;4ac}{4a},\;\infty \right )}\) or \(\mathbf{\left [ -\;\frac{D}{4a},\;\infty \right )}\)

Therefore, the Minimum value of f(x) is:

i.e. \(\mathbf{-\;\frac{144\;-\;44}{4}}\) at x = \(\mathbf{-\;\frac{-12}{2}}\) = – 25 at x = 6.

The Maximum value of f(x) is infinity.

Therefore, the Range of given quadratic equation is [- 25, ∞).

Example 2: Find the minimum value of equation \(\mathbf{\frac{2x\;-\;5}{2x^{2}\;+\;3\;x\;+\;6}}\) ?

Solution:

From the given quadratic expression f (x), a > 0. Therefore f (x) will have minimum value at x = \(\mathbf {\frac {-\;b}{2a}}\).

i.e. at x = \(\mathbf {\frac {-3}{4}}\) = \(\mathbf {\frac {-3}{4}}\).

And, the minimum value of quadratic equation f (x) = \(\mathbf {\frac {-\;D}{4a}}\) = \(\mathbf {-\;\frac {9\;-\;48}{8}}\) = \(\mathbf {\frac {39}{8}}\)

Therefore, the minimum value of equation \(\mathbf{\frac{2x\;-\;5}{2x^{2}\;+\;3\;x\;+\;6}}\) = \(\mathbf {\frac {39}{8}}\)

Example 3: For what values of ‘m’, the quadratic Expression \(x^{2} + 2 (m + 1)x + 9m – 5 = 0\) will have only negative roots.

Solution:

Consider the given quadratic Expression f (x) = \(x^{2} + 2 (m + 1)x + 9m – 5 \)

Here a = 1, b = 2 (m + 1), and c = 9m – 5.

Therefore, D = \(b^{2} – 4ac\) = \(4 (m + 1)^{2} – 4 (9m – 5) = 4m^{2} + 24 – 28 m . . . . . . . . . (1)\)

α + β = – 2 (m + 1) and αβ = 9m – 5 . . . . . . . . . . . . . . (2)

Now, both the roots of a quadratic Expression are negative if, \(b^{2} – 4ac\) = (D) ≥ 0, α + β = \(\mathbf{\frac{-b}{a}}\) < 0, and αβ = \(\mathbf{\frac{c}{a}}\) < 0

Therefore, \(4m^{2} + 24 – 28 m ≥ 0\) [From equation (1)]

i.e. \(m^{2} – 7m + 6 ≥ 0\)

Or, (m – 1) (m – 6) ≥ 0

Therefore, m ≤ 1 or m ≥ 6 . . . . . . . . . . (3)

Now, α + β < 0

i.e. – 2(m + 1) < 0 [From equation (2)]

Or, m > -1 . . . . . . . (4)

And, α β < 0

i.e. 9m – 5 < 0

Or, m < \(\mathbf{\frac{5}{9}}\) . . . . . . (5)

From Equations (3), (4), and (5) we can conclude that the given quadratic equation will have only negative roots if m ≥ 6.

Example 4: Find the range of function f(x) = \(\mathbf{\frac{x\;+\;2}{2x^{2}\;+\;3x\;+\;6}}\), if x is real.

Solution:

Let y = \(\mathbf{\frac{x\;+\;2}{2x^{2}\;+\;3x\;+\;6}}\)

Then, \(2 y x^{2} + 3xy +6y = x + 2\)

Or, \(2yx^{2} + (3y – 1)x + 6y – 2 = 0\)

Now, Discriminant (D) = \(b^{2} – 4ac\) ≥ 0 [because x is real]

i.e. \((3y – 1)^{2} – 4 (2y) (6y – 2) ≥ 0\)

Or, \(9y^{2} + 1 – 6y – 48y^{2} + 16y ≥ 0\)

Or, \(- 39 y^{2} + 10y + 1 ≥ 0\)

Or, \(39 y^{2} – 10y – 1 \leq 0\)

Or, \(39 y^{2} – 13y + 3y – 1 \leq 0\)

Or, \(13y (3y – 1) + 1 (3y – 1) ≤ 0\)

Therefore, the range of function f(x) = \(\mathbf{\frac{x\;+\;2}{2x^{2}\;+\;3x\;+\;6}}\):

\(\mathbf{y\;\in \;\left [ \;-\;\frac{1}{13},\;\frac{1}{3} \;\right ]}\)


Practise This Question

The Undertaker brags about stopping a moving car with bare hands. To prove himself he stands infront of a moving car coming towards him and surprisingly stops the car in 10 seconds after initial contact.The work done by him in this duration will be