The equation of the path traversed by the ball headed by the footballer is $y = ax^2 + bx + c$; (where $0 \le x \le 14$ and $a, b, c \in R$ and $a \ne 0$) with respect to a XY-coordinate system in the vertical plane. The ball passes through the points $(2, 15)$, $(4, 25)$ and $(14, 15)$. Determine the values of $a$, $b$ and $c$ by solving the system of linear equations in $a$, $b$ and $c$, using matrix method. Also find the equation of the path traversed by the ball.

The equation of the path traversed by the ball headed by the footballer is y = ax² + bx + c; (where 0 ≤ x ≤ 14

Question

y = ax^2 + bx + c

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✓ Free question

Concept understanding — Quadratic Extrema

Quadratic Extrema: From Intuition to Precision

Toss a ball straight up: it rises, slows, stops for an instant at the top, then falls. Plot its height against time and you get a parabola with exactly one turning point — a peak (maximum) or a valley (minimum). That single highest or lowest point is what quadratic extrema are about.

The Intuition First

A quadratic is $f(x) = ax^2 + bx + c$ , with $a \neq 0$ ; its graph is a parabola.

If $a > 0$ , it opens upward (a U) and has a minimum at the bottom.
If $a < 0$ , it opens downward and has a maximum at the top.

The turning point is the vertex. Every quadratic has exactly one vertex — that's the extremum.

Note

Unlike cubic or higher-degree polynomials, a quadratic never has both a maximum and a minimum. It has one or the other.

The Precise Statement

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

Vertex (extremum) at

x = -\frac{b}{2a}

Extremum value

f\left(-\frac{b}{2a}\right) = c - \frac{b^2}{4a}

Nature: $a > 0$ → minimum; $a < 0$ → maximum.

\text{Vertex} = \left( -\frac{b}{2a},\; c - \frac{b^2}{4a} \right)

Why That $x$ ? A Quick Derivation

Complete the square:

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

The squared term is always $\geq 0$ . When $a > 0$ , $f(x)$ is smallest when the square is zero — at $x = -\frac{b}{2a}$ . When $a < 0$ , the largest value occurs at the same $x$ .

Tip

The vertex's $x$ -coordinate is also the average of the two roots (if they exist): $x = \frac{\text{root}_1 + \text{root}_2}{2}$ .

Common Mistake to Avoid

Watch out

Don't confuse the sign of $a$ with the sign of the extremum value. With $a > 0$ you always have a minimum, but that minimum could be positive, negative, or zero. The shape tells you max vs min, not the number itself.

Example

Find the extremum of $f(x) = 2x^2 - 8x + 5$ .

Here $a = 2 > 0$ , so it's a minimum.

$x = -\frac{-8}{2 \cdot 2} = 2$ .

$f(2) = 8 - 16 + 5 = -3$ .

So the minimum is at $(2, -3)$ .

The Big Picture

Quadratic extrema are the simplest non-trivial optimization problem in algebra, appearing in projectile motion, profit maximization, area problems, and least-squares regression. One formula, one turning point, and the sign of $a$ decides peak or valley.

Finding the vertex of a quadratic function via -b/2a is foundational algebra from the NCERT Class 11 units on quadratic expressions, and it reappears as a special case of the general maxima-minima methods in the NCERT Class 12 Application of Derivatives chapter. Students searching 'maximum and minimum value of quadratic function' or 'vertex formula class 11 maths examples' will recognize this completing-the-square derivation as exactly the shortcut those questions expect.

Substituting the three points gives a $3\times 3$ linear system; solving by the matrix method yields $a = -\tfrac{1}{2},\ b = 8,\ c = 1$ , so the path is $y = -\tfrac{1}{2}x^2 + 8x + 1$ .

Form the equations. With $y = ax^2 + bx + c$ passing through $(2,15)$ , $(4,25)$ , $(14,15)$ :

4a + 2b + c = 15, \qquad 16a + 4b + c = 25, \qquad 196a + 14b + c = 15.

Write in matrix form $AX = B$ :

A = \begin{bmatrix} 4 & 2 & 1 \\ 16 & 4 & 1 \\ 196 & 14 & 1 \end{bmatrix},\quad X = \begin{bmatrix} a \\ b \\ c \end{bmatrix},\quad B = \begin{bmatrix} 15 \\ 25 \\ 15 \end{bmatrix}.

Determinant of $A$ :

|A| = 4(4 - 14) - 2(16 - 196) + 1(224 - 784) = -40 + 360 - 560 = -240 \ne 0,

so $A^{-1}$ exists and $X = A^{-1}B$ has a unique solution.

Solve (eliminate $c$ ). Subtracting equation 1 from equation 2, and equation 2 from equation 3:

12a + 2b = 10 \ \Rightarrow\ 6a + b = 5, \qquad 180a + 10b = -10 \ \Rightarrow\ 18a + b = -1.

Subtracting these: $12a = -6 \Rightarrow a = -\tfrac{1}{2}$ . Then $b = 5 - 6a = 5 + 3 = 8$ , and from equation 1, $c = 15 - 4a - 2b = 15 + 2 - 16 = 1$ .

Check: at $(14,15)$ : $-\tfrac{1}{2}(196) + 8(14) + 1 = -98 + 112 + 1 = 15.$ ✓

✓Final answer

a = -\dfrac{1}{2},\ b = 8,\ c = 1

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 mark
$f$
›Reveal solutionSolution

$f(x)=x^2\ge0$ , with minimum $0$ at $x=0$ .

For all real $x$ , $x^2\ge0$ , and equality holds at $x=0$ . So the least value of $f$ is $0$ .

✓Final answer
$0$
CBSE 2024Set A1 mark
$f$
›Reveal solutionSolution

$f(x)=x^2$ attains its minimum value $0$ at $x=0$ .

Since $x^2\ge 0$ for every real $x$ , with equality only at $x=0$ , the function $f(x)=x^2$ has a global minimum value of $0$ at $x=0$ and no maximum value on $\mathbb{R}$ (it grows without bound as $x\to\pm\infty$ ).

✓Final answer
$0$
CBSE 2022Set ANNUAL1 mark
$f(x)=x^2-4x+5$
›Reveal solutionSolution

Set $f'(x)=0$ to find the critical point, then use the sign of $f''(x)$ to classify it as max/min.

$f(x)=x^2-4x+5 \Rightarrow f'(x)=2x-4$ .

Critical point: $2x-4=0\Rightarrow x=2$ .

$f''(x)=2>0$ , so $x=2$ gives a minimum value, $f(2)=4-8+5=1$ .

✓Final answer
$x=2$
CBSE 2020Set 65/3/11 markMCQ
$y = -x^3 + 3x^2 + 12x - 5$
›Reveal solutionSolution
The slope of a curve is its derivative; to maximize the slope, we find the critical points of the derivative by setting the second derivative to zero. The maximum slope is $15$ .

The slope of a curve at any point is given by its first derivative. So we're really being asked: what is the maximum value of $\frac{dy}{dx}$ ?

To maximize a function, we treat it like any other optimization problem. Once we have an expression for the slope, we find where that function reaches its maximum by using calculus again—taking its derivative (the second derivative of the original function) and finding critical points.

Let me work through this systematically.

Find the slope function

Differentiate $y = -x^3 + 3x^2 + 12x - 5$ :

$\frac{dy}{dx} = -3x^2 + 6x + 12$
This is the slope at any point $x$ on the curve.

Set up the maximization problem

We want to maximize $f(x) = -3x^2 + 6x + 12$ . Notice this is a quadratic function in $x$ with a negative leading coefficient, so it opens downward and has a maximum at its vertex.

Find the critical point

Take the derivative of the slope function:

$\frac{d^2y}{dx^2} = -6x + 6$
Set this equal to zero:
$-6x + 6 = 0$ $x = 1$

Verify it's a maximum

The third derivative is $\frac{d^3y}{dx^3} = -6 < 0$ , confirming that the second derivative changes from positive to negative at $x = 1$ , so the first derivative (slope) has a maximum there.

Alternatively, since $\frac{dy}{dx} = -3x^2 + 6x + 12$ is a downward-opening parabola, its vertex is indeed a maximum.

Calculate the maximum slope

Substitute $x = 1$ into the slope function:

$\frac{dy}{dx}\bigg|_{x=1} = -3(1)^2 + 6(1) + 12 = -3 + 6 + 12 = 15$
Tip

For any quadratic $ax^2 + bx + c$ with $a < 0$ , the maximum occurs at $x = -\frac{b}{2a}$ . Here, with $-3x^2 + 6x + 12$ , that gives $x = -\frac{6}{2(-3)} = 1$ , leading directly to the maximum value.

✓Final answer
$\boxed{15}$

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