How to Find Maximum and Minimum Values of a Quadratic Polynomial

A quadratic polynomial is a real-valued function of the form $f(x) = ax^2 + bx + c$, where $a \neq 0$ and $a, b, c \in \mathbb{R}$. The maximum and minimum values of such a polynomial are determined by the sign and magnitude of the leading coefficient $a$ and the properties of its graph.

Formal Structure and Vertex of a Quadratic Polynomial

The standard form of a quadratic polynomial is $f(x) = ax^2 + bx + c$ with $a \neq 0$. The graph of $f(x)$ is a parabola. The orientation of the parabola depends on the sign of $a$: if $a > 0$, it opens upwards; if $a < 0$, it opens downwards.

For any quadratic polynomial, the parabola attains its unique extremum (maximum or minimum) at the vertex. The $x$-coordinate of the vertex is given by $x_v = -\frac{b}{2a}$. This point is found by differentiating $f(x)$, setting $f'(x)=0$, and solving for $x$.

Substituting $x_v$ back into $f(x)$ provides the corresponding $y$-coordinate (value of the extremum):

$f(x_v) = a\left(-\dfrac{b}{2a}\right)^2 + b\left(-\dfrac{b}{2a}\right) + c$

Expanding, $f(x_v) = a\dfrac{b^2}{4a^2} - \dfrac{b^2}{2a} + c$

$= \dfrac{b^2}{4a} - \dfrac{b^2}{2a} + c$

Writing with a common denominator, $= \dfrac{b^2 - 2b^2}{4a} + c = \dfrac{-b^2}{4a} + c$

The extremum value is $f(x_v) = c - \dfrac{b^2}{4a}$.

Maximum Value of a Quadratic Polynomial: Criterion and Precise Value

If $a < 0$, the parabola opens downwards and the quadratic polynomial attains an absolute maximum at its vertex. The function increases for $x < x_v$, is stationary at $x = x_v$, and decreases for $x > x_v$.

If $a < 0$, the maximum value of $f(x) = ax^2 + bx + c$ is $c - \dfrac{b^2}{4a}$, achieved at $x = -\dfrac{b}{2a}$.

For $a > 0$, the function has no finite maximum as $f(x) \to +\infty$ as $|x| \to \infty$. The value $c - \dfrac{b^2}{4a}$ is the minimum, not the maximum, in this case.

Minimum Value of a Quadratic Polynomial: Criterion and Precise Value

If $a > 0$, the parabola opens upwards and $f(x)$ attains an absolute minimum at its vertex. The function decreases for $x < x_v$, is stationary at $x = x_v$, and increases for $x > x_v$.

If $a > 0$, the minimum value of $f(x) = ax^2 + bx + c$ is $c - \dfrac{b^2}{4a}$, attained at $x = -\dfrac{b}{2a}$.

For $a < 0$, the minimum value is not finite, since $f(x) \to -\infty$ as $|x| \to \infty$, so the minimum does not exist in $\mathbb{R}$.

Explicit Derivation of the Extremum Value Formula

Consider $f(x) = ax^2 + bx + c$. Complete the square to determine the extremum value.

Rewrite $f(x)$ as:

$f(x) = a(x^2 + \dfrac{b}{a}x) + c$

$= a(x^2 + \dfrac{b}{a}x + \left(\dfrac{b}{2a}\right)^2) - a\left(\dfrac{b}{2a}\right)^2 + c$

$= a\left(x + \dfrac{b}{2a}\right)^2 - a\left(\dfrac{b^2}{4a^2}\right) + c$

$= a\left(x + \dfrac{b}{2a}\right)^2 + c - \dfrac{b^2}{4a}$

The minimum or maximum occurs when $x + \dfrac{b}{2a} = 0$, i.e., $x = -\dfrac{b}{2a}$.

Substituting, $f\left(-\dfrac{b}{2a}\right) = a(0)^2 + c - \dfrac{b^2}{4a} = c - \dfrac{b^2}{4a}$.

The direction of extremum depends on the sign of $a$: for $a > 0$ it is a minimum, for $a < 0$ it is a maximum.

Maximum and Minimum for Quadratic Polynomials on Intervals

If $f(x)$ is considered on a closed interval $[p, q]$, the absolute maximum and minimum must occur at either the vertex $x = -\dfrac{b}{2a}$, provided it lies in $[p, q]$, or at an endpoint. Evaluate $f(x)$ at $x = p$, $x = q$, and $x = -\dfrac{b}{2a}$ (if within $[p, q]$); the largest yields the maximum, the smallest yields the minimum.

Nature of Roots and Connection to Extremum

The sign of the discriminant $D = b^2 - 4ac$ determines the nature of roots of $f(x) = 0$, but it does not affect the maximum or minimum value of the quadratic polynomial on $\mathbb{R}$.

Examples: Calculation of Maximum or Minimum Values

Example. Given $f(x) = 2x^2 + 7x + 5$.

Here $a = 2 > 0$; the parabola opens upwards. Thus, $f(x)$ has a minimum at the vertex.

$x_v = -\dfrac{b}{2a} = -\dfrac{7}{2 \times 2} = -\dfrac{7}{4}$

$f(x_v) = 2\left(-\dfrac{7}{4}\right)^2 + 7\left(-\dfrac{7}{4}\right) + 5$

$= 2 \cdot \dfrac{49}{16} - \dfrac{49}{4} + 5$

$= \dfrac{98}{16} - \dfrac{196}{16} + \dfrac{80}{16}$

$= \dfrac{98 - 196 + 80}{16}$

$= \dfrac{-18}{16} = -\dfrac{9}{8}$

Thus, the minimum value is $-\dfrac{9}{8}$ at $x = -\dfrac{7}{4}$.

Example. Let $f(x) = -5x^2 + 30x + 200$. Here $a = -5 < 0$; the parabola opens downwards. The function has a maximum at $x_v = -\dfrac{b}{2a} = -\dfrac{30}{2 \times (-5)} = 3$.

$f(3) = -5(3^2) + 30 \cdot 3 + 200$

$= -45 + 90 + 200$

$= 245$

Thus, the maximum value is $245$, attained at $x = 3$.

Exam Notes and Frequently Misapplied Cases

Exam Tip. To distinguish between maximum and minimum, check the sign of the leading coefficient $a$. Positive $a$ implies a minimum; negative $a$ implies a maximum.

A common error is to use $c$ as the extremum value regardless of $b$ and $a$; always use $c - \dfrac{b^2}{4a}$.

Summary Results for Maximum and Minimum Values

Result. For $f(x) = ax^2 + bx + c$, $a \neq 0$:

If $a > 0$, the minimum value is $c - \dfrac{b^2}{4a}$ at $x = -\dfrac{b}{2a}$; no finite maximum.

If $a < 0$, the maximum value is $c - \dfrac{b^2}{4a}$ at $x = -\dfrac{b}{2a}$; no finite minimum.

For further study,  refer to Maximum And Minimum Value Of Trigonometric Functions.

For graphical understanding of extremum in quadratic polynomials, consult Graphs Of Sine And Cosine Function.