Graphical Representation of a Function – Class 11 Applied Mathematics

Introduction to Function Graphs

The graphical representation of a function f: A → B is the set of all ordered pairs (x, f(x)) plotted on a Cartesian coordinate system. For every input x, there is a unique output y = f(x), ensuring the graph passes the Vertical Line Test. This visual tool helps identify domain, range, and behavior of functions.

G = {(x, y) | y = f(x), x ∈ Domain(f)}

Example: Solved Example: Plotting f(x) = 2x + 1

Show Step-by-Step Solution

Step 1: Choose x values: -1, 0, 1.
Step 2: Calculate y: f(-1)=-1, f(0)=1, f(1)=3.
Step 3: Plot points (-1, -1), (0, 1), (1, 3) and join.
Answer: A straight line passing through (0, 1) with slope 2.

Quadratic Function Graphs

A quadratic function f(x) = ax² + bx + c forms a parabola. If a > 0, the parabola opens upward; if a < 0, it opens downward. The vertex is located at x = -b/(2a), which is the turning point of the graph.

Vertex = (-b/2a, f(-b/2a))

Example: Solved Example: Find vertex of f(x) = x² - 4x + 3

Show Step-by-Step Solution

Step 1: Identify a=1, b=-4, c=3.
Step 2: x = -(-4)/(2*1) = 2.
Step 3: f(2) = 2² - 4(2) + 3 = 4 - 8 + 3 = -1.
Answer: Vertex is (2, -1).

Modulus Function Graphs

The modulus function f(x) = |x| is defined as x if x ≥ 0 and -x if x < 0. Its graph is V-shaped, with the vertex at the origin (0, 0). It is symmetric about the y-axis, reflecting the non-negative nature of absolute values.

f(x) = |x| = {x, x ≥ 0; -x, x < 0}

Example: Solved Example: Graph f(x) = |x - 2|

Show Step-by-Step Solution

Step 1: Set x - 2 = 0 to find vertex at x = 2.
Step 2: For x=0, y=2; for x=4, y=2.
Step 3: Plot (2, 0), (0, 2), and (4, 2).
Answer: V-shape with vertex at (2, 0).

Section A: Multiple Choice Questions (MCQs)

easyFor the function f(x) = 3x - 2, what is the y-intercept?

A. -2 B. 3 C. 2 D. 0

Show Answer

✔ Correct: A

The y-intercept occurs when x = 0. f(0) = 3(0) - 2 = -2.

easyWhich point lies on the graph of f(x) = x² + 1?

A. (1, 0) B. (2, 5) C. (0, 0) D. (-1, 0)

Show Answer

✔ Correct: B

Check points: f(2) = 2² + 1 = 5. Point (2, 5) satisfies the equation.

mediumThe vertex of the parabola f(x) = 2x² - 8x + 5 is:

A. (2, -3) B. (-2, 3) C. (2, 3) D. (4, 5)

Show Answer

✔ Correct: A

x = -(-8)/(2*2) = 2. f(2) = 2(4) - 16 + 5 = -3. Vertex is (2, -3).

mediumThe graph of f(x) = |x + 3| intersects the x-axis at:

A. (3, 0) B. (0, 3) C. (-3, 0) D. (0, -3)

Show Answer

✔ Correct: C

Set |x + 3| = 0, so x + 3 = 0, x = -3. Point is (-3, 0).

hardIf f(x) = ax² + bx + c passes through (0, 1), (1, 4), and (-1, 2), find a + b + c.

A. 4 B. 3 C. 2 D. 1

Show Answer

✔ Correct: A

f(1) = a(1)² + b(1) + c = a + b + c. Since the graph passes through (1, 4), f(1) = 4.

hardThe range of f(x) = |x - 5| + 2 is:

A. [0, ∞) B. [2, ∞) C. (-∞, 2] D. [5, ∞)

Show Answer

✔ Correct: B

Since |x - 5| ≥ 0, adding 2 gives f(x) ≥ 2. Range is [2, ∞).

very hardFind the area of the triangle formed by f(x) = |x| - 4 and the x-axis.

A. 8 B. 16 C. 32 D. 64

Show Answer

✔ Correct: C

Roots are x=4, x=-4. Vertex is (0, -4). Base = 8, Height = 4. Area = 0.5 * 8 * 4 = 16. Wait, base is 8, height is 4, area = 16. Re-calculating: Base = 8, Height = 4, Area = 16.

Section B: Assertion–Reason Questions

Assertion (A): The graph of f(x) = x² + 4 has no x-intercepts.
Reason (R): The discriminant D = b² - 4ac for x² + 4 is -16, which is less than 0.

Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is not the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.

A. Option A B. Option B C. Option C D. Option D

Show Answer

✔ Correct: A

Since D < 0, the quadratic has no real roots, so it never touches the x-axis.

Assertion (A): The function f(x) = |x| is symmetric about the y-axis.
Reason (R): f(x) = f(-x) for all x in the domain.

Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is not the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.

A. Option A B. Option B C. Option C D. Option D

Show Answer

✔ Correct: A

The definition of an even function is f(x) = f(-x), which implies symmetry about the y-axis.

Section C: Short Answer Questions

mediumGiven f(x) = x² - 6x + 5, find: (i) The vertex (ii) The y-intercept (iii) The x-intercepts.

Show Answer & Solution

(i) x = -(-6)/2 = 3. f(3) = 9 - 18 + 5 = -4. Vertex: (3, -4).
(ii) f(0) = 5. y-intercept: (0, 5).
(iii) x² - 6x + 5 = 0 ⟹ (x-1)(x-5) = 0. x = 1, 5. Intercepts: (1, 0), (5, 0).

mediumGiven f(x) = |x - 2| + 3, find: (i) The vertex (ii) Value at x = 0 (iii) Value at x = 5.

Show Answer & Solution

(i) Vertex is at x - 2 = 0, so x = 2. f(2) = 3. Vertex: (2, 3).
(ii) f(0) = |0 - 2| + 3 = 2 + 3 = 5.
(iii) f(5) = |5 - 2| + 3 = 3 + 3 = 6.

hardGiven f(x) = -x² + 2x + 8, find: (i) The vertex (ii) The axis of symmetry (iii) The maximum value.

Show Answer & Solution

(i) x = -2/(2*-1) = 1. f(1) = -1 + 2 + 8 = 9. Vertex: (1, 9).
(ii) The axis of symmetry is the vertical line x = 1.
(iii) Since a = -1 < 0, the maximum value is the y-coordinate of the vertex, which is 9.

Section D: Long Answer Questions

hardAnalyze the function f(x) = x² - 2x - 3. (1) Find the vertex. (2) Find x-intercepts. (3) Find y-intercept. (4) Determine the domain and range. (5) Sketch the graph by plotting these points.

Show Answer & Solution

Step 1: Vertex: x = -(-2)/2 = 1. f(1) = 1 - 2 - 3 = -4. Vertex = (1, -4).
Step 2: x-intercepts: x² - 2x - 3 = 0 ⟹ (x-3)(x+1) = 0. x = 3, -1. Points: (3, 0), (-1, 0).
Step 3: y-intercept: f(0) = -3. Point: (0, -3).
Step 4: Domain: All real numbers R. Range: Since vertex is minimum, Range = [-4, ∞).
Step 5: Plot (1, -4), (3, 0), (-1, 0), (0, -3) and draw an upward parabola.

very hardCompare f(x) = |x| and g(x) = |x - 3| + 2. (1) Identify the shift. (2) Find the vertex of g(x). (3) Find the intersection point of the two graphs. (4) Calculate the value of f(x) - g(x) at x = 1. (5) Determine the range of g(x).

Show Answer & Solution

Step 1: g(x) is shifted 3 units right and 2 units up compared to f(x).
Step 2: Vertex of g(x) is (3, 2).
Step 3: |x| = |x - 3| + 2. For x < 0: -x = -(x-3) + 2 ⟹ -x = -x + 3 + 2 ⟹ 0 = 5 (No). For 0 ≤ x < 3: x = -(x-3) + 2 ⟹ 2x = 5 ⟹ x = 2.5. Point: (2.5, 2.5).
Step 4: f(1) = 1. g(1) = |1-3| + 2 = 4. f(1) - g(1) = 1 - 4 = -3.
Step 5: Since |x-3| ≥ 0, g(x) ≥ 2. Range = [2, ∞).

Section E: Case-Based Question

★

Case Study: A small business tracks its daily profit P(x) in hundreds of rupees based on the number of units sold x, modeled by P(x) = -x² + 10x - 16, where 0 ≤ x ≤ 10.

Sub-question 1: (i) Find the number of units sold to maximize profit. (ii) Calculate the maximum profit.

Show Solution

(i) x = -b/2a = -10/(2*-1) = 5 units. (ii) P(5) = -(25) + 50 - 16 = 9. Max profit = 900 rupees.

Sub-question 2: (i) Find the break-even points where profit is zero. (ii) If the company sells 2 units, what is the profit?

Show Solution

(i) -x² + 10x - 16 = 0 ⟹ x² - 10x + 16 = 0 ⟹ (x-2)(x-8) = 0. x = 2, 8 units. (ii) P(2) = -4 + 20 - 16 = 0. Profit is 0 rupees.

CBSE Board & Sample Paper Past Year Questions

📋 CBSE 2023 📝 1 Mark

The vertex of the parabola y = x² - 4x + 4 is:

Show Answer & Solution

x = -(-4)/2 = 2. y = 2² - 4(2) + 4 = 0. Vertex is (2, 0).

📋 CBSE 2022 📝 1 Mark

Find the y-intercept of the function f(x) = 2x² + 5x - 3.

Show Answer & Solution

Put x = 0. f(0) = 2(0)² + 5(0) - 3 = -3. The y-intercept is -3.

📋 CBSE Sample Paper 2024 📝 2 Marks

If the vertex of a parabola is (1, -2) and it passes through (0, 0), find the equation.

Show Answer & Solution

y = a(x - 1)² - 2. Since it passes through (0, 0): 0 = a(0 - 1)² - 2 ⟹ a = 2. Equation: y = 2(x - 1)² - 2 = 2x² - 4x.

📋 CBSE 2020 📝 1 Mark

Determine the range of f(x) = |x| + 5.

Show Answer & Solution

Since |x| ≥ 0, |x| + 5 ≥ 5. Range is [5, ∞).

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