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Unit 3: Theory of Quadratic Functions — Exercise 3 1

11th Class Mathematics · Unit 3: Theory of Quadratic Functions

1.Find the maximum or minimum value of the following quadratic function by completing the square: (i) f(x) = x² + 6x + 13 (ii) f(x) = x² + 4x (iii) f(x) = -x² + 8x + 13 (iv) f(x) = -x² - 3x - 5 (v) f(x) = 3x² + 6x - 13 (vi) f(x) = -2x² - x + 21

(i) f(x) = x² + 6x + 9 + 4

f(x) = (x + 3)² + 4

Comparing with standard form f(x) = a(x - h)² + k, here a = 1 > 0

Result

Minimum value 4 at x = -3

(ii) f(x) = x² + 4x + 4 - 4

f(x) = (x + 2)² - 4

Comparing with standard form f(x) = a(x - h)² + k, here a = 1 > 0

Minimum value -4 at x = -2

(iii) f(x) = -x² + 8x + 13 + 16 - 16

f(x) = -x² + 8x - 16 + 29

f(x) = -(x² - 2(4)x + 16) + 29

f(x) = -(x - 4)² + 29

Comparing with standard form f(x) = a(x - h)² + k, here a = -1 < 0

Maximum value 29 at x = 4

(iv) f(x) = -x² - 3x - 5 + 94 - 94

f(x) = -x² - 3x - 94 - 114

f(x) = -(x² + 3x + 94) - 114

f(x) = -(x + 32)² - 114

Comparing with standard form f(x) = a(x - h)² + k, here a = -1 < 0

Maximum value -114 at x = -32

(v) f(x) = 3(x² + 2x) - 13

f(x) = 3(x² + 2x + 1 - 1) - 13

f(x) = 3[(x + 1)² - 1] - 13

f(x) = 3(x + 1)² - 3 - 13

f(x) = 3(x + 1)² - 16

Comparing with standard form f(x) = a(x - h)² + k, here a = 3 > 0

Minimum value -16 at x = -1

(vi) f(x) = -2(x² + x2) + 21

f(x) = -2(x² + x2 + 116 - 116) + 21

f(x) = -2[(x² + x2 + (14)²) - 116] + 21

f(x) = -2[(x + 14)² - 116] + 21

f(x) = -2(x + 14)² + 18 + 21

f(x) = -2(x + 14)² + 1698

Comparing with standard form f(x) = a(x - h)² + k, here a = -2 < 0

Maximum value 1698 at x = -14

2.Find the maximum or minimum point by sketching the following quadratic function. Also find its domain and range: (i) f(x) = x² - 4x (ii) f(x) = x² - 5x + 6 (iii) f(x) = -x² + 2x - 8 (iv) f(x) = x² - 4x + 4 (v) f(x) = x² + 2x - 8.3 (vi) f(x) = 6 - x - x²

Given

(i) Function f(x) = x² - 4x

y-intercept y = (0)² - 4(0) = 0

x-intercepts x² - 4x = 0 ⇒ x(x - 4) = 0 ⇒ x = 0, x = 4

Vertex x-coordinate h = -b2a = --42(1) = 2

Vertex y-coordinate k = (2)² - 4(2) = 4 - 8 = -4

Result

Vertex (minimum point) (2, -4)

Domain (-infty, infty)

Range [-4, infty)

Given

(ii) Function f(x) = x² - 5x + 6

y-intercept y = (0)² - 5(0) + 6 = 6

x-intercepts from factoring x² - 5x + 6 = 0 ⇒ x² - 3x - 2x + 6 = 0 ⇒ (x - 3)(x - 2) = 0 ⇒ x = 3, x = 2

Vertex x-coordinate h = -b2a = --52(1) = 52

Vertex y-coordinate k = (52)² - 5(52) + 6 = 254 - 252 + 6 = 25 - 50 + 244 = -14

Result

Vertex (minimum point) (52, -14)

Domain (-infty, infty)

Range [-14, infty)

Given

(iii) Function f(x) = -x² + 2x - 8

y-intercept y = -(0)² + 2(0) - 8 = -8

x-intercepts -x² + 2x - 8 = 0 ⇒ x² - 2x + 8 = 0

Discriminant Δ = (-2)² - 4(1)(8) = 4 - 32 = -28 < 0

Note As roots are complex, parabola does not cross x-axis

Vertex x-coordinate h = -b2a = -22(-1) = 1

Vertex y-coordinate k = -(1)² + 2(1) - 8 = -1 + 2 - 8 = -7

Result

Vertex (maximum point) (1, -7)

Domain (-infty, infty)

Range (-infty, -7]

Given

(iv) Function f(x) = x² - 4x + 4

y-intercept y = (0)² - 4(0) + 4 = 4

x-intercepts from quadratic formula x = frac{-(-4) ± sqrt{(-4)² - 4(1)(4)}{2(1)} = frac{4 ± sqrt{16 - 16}{2} = 4 ± 02 ⇒ x = 2

Vertex x-coordinate h = -b2a = --42(1) = 2

Vertex y-coordinate k = (2)² - 4(2) + 4 = 4 - 8 + 4 = 0

Result

Vertex (minimum point) (2, 0)

Domain (-infty, infty)

Range [0, infty)

Given

(v) Function f(x) = x² + 2x - 8.3

y-intercept y = (0)² + 2(0) - 8.3 = -8.3

x-intercepts x² + 2x - 8.3 = 0 ⇒ x² + 2x - 8310 = 0 ⇒ 10x² + 20x - 83 = 0

Using quadratic formula x = frac{-20 ± sqrt{400 + 3320}{20} = frac{-20 ± sqrt{3720}{20} = frac{-20 ± 2sqrt{930}{20} = frac{-10 ± sqrt{930}{10}

Numerical values x = 2.049, x = -4.049

Vertex x-coordinate h = -b2a = -22(1) = -1

Vertex y-coordinate k = (-1)² + 2(-1) - 8.3 = 1 - 2 - 8.3 = -9.3

Result

Vertex (minimum point) (-1, -9.3)

Domain (-infty, infty)

Range [-9.3, infty)

Given

(vi) Function f(x) = 6 - x - x²

y-intercept y = 6 - 0 - (0)² = 6

x-intercepts 6 - x - x² = 0 ⇒ x² + x - 6 = 0 ⇒ (x + 3)(x - 2) = 0 ⇒ x = -3, x = 2

Rewrite as standard form f(x) = -x² - x + 6

Vertex x-coordinate h = -b2a = --12(-1) = -12

Vertex y-coordinate k = 6 - (-12) - (-12)² = 6 + 12 - 14 = 254

Result

Vertex (maximum point) (-12, 254)

Domain (-infty, infty)

Range (-infty, 254]

3.Find the inverse of the following quadratic function. Also find its domain and range: (i) f(x) = x² - 3, x ≤ 0 (ii) f(x) = x² + 6x + 4, x < -3 (iii) f(x) = 2x² - 8x + 11, x ≥ 2 (iv) f(x) = 3x² - 2x + 6, x ≥ 5 (v) f(x) = 2(x - 3)² + 1, x ≥ 3 (vi) f(x) = -3(x + 4)² - 5, x < -4

Given

(i) Function f(x) = x² - 3, quad x le 0

Let y = x² - 3

Interchange x and y x = y² - 3

Solve for y y² = x + 3 ⇒ y = ± sqrt{x + 3}

Domain of original (-infty, 0]

For range of original x le 0 ⇒ x² ge 0 ⇒ x² - 3 ge -3

Range of original f(x) ge -3

Therefore Range of f(x) = [-3, infty)

Domain of inverse [-3, infty)

Range of inverse (-infty, 0]

Testing at x = 0 f^-1(0) = sqrt{0 + 3} = sqrt{3} or f^-1(0) = -sqrt{3}

Since original domain is x ≤ 0 Only -sqrt{3} lies in the range (-infty, 0]

Result

Inverse f^-1(x) = -sqrt{x + 3}

Given

(ii) Function f(x) = x² + 6x + 4, quad x < -3

Let y = x² + 6x + 4

Interchange x and y x = y² + 6y + 4

Rearrange y² + 6y + 4 - x = 0

Using quadratic formula y = frac{-6 ± sqrt{36 - 4(4 - x)}{2} = frac{-6 ± sqrt{36 - 16 + 4x}{2} = frac{-6 ± sqrt{20 + 4x}{2}

Simplify y = frac{-6 ± 2sqrt{5 + x}{2} = -3 ± sqrt{5 + x}

Domain of original (-infty, -3)

For range of original f(x) = (x + 3)² - 5, quad x < -3

Since x < -3 x + 3 < 0 ⇒ (x + 3)² > 0 ⇒ f(x) > -5

Range of original (-5, infty)

Domain of inverse (-5, infty)

Range of inverse (-infty, -3)

Testing at x = -4 f^-1(-4) = -3 + sqrt{5 - 4} = -3 + 1 = -2 or -3 - 1 = -4

Only -4 is in range Therefore discard positive component

Result

Inverse f^-1(x) = -3 - sqrt{5 + x}

Given

(iii) Function f(x) = 2x² - 8x + 11, quad x ge 2

Let y = 2x² - 8x + 11

Interchange x and y x = 2y² - 8y + 11

Rearrange 2y² - 8y + 11 - x = 0

Using quadratic formula y = frac{-(-8) ± sqrt{64 - 4(2)(11 - x)}{2(2)} = frac{8 ± sqrt{64 - 88 + 8x}{4} = frac{8 ± sqrt{8x - 24}{4}

Simplify y = frac{8 ± 2sqrt{2x - 6}{4} = frac{4 ± sqrt{2x - 6}{2}

Domain of original [2, infty)

By completing square f(x) = 2(x² - 4x + 4 - 4) + 11 = 2(x - 2)² + 3, quad x ge 2

For range x ge 2 ⇒ (x - 2)² ge 0 ⇒ 2(x - 2)² + 3 ge 3

Range of original [3, infty)

Domain of inverse [3, infty)

Range of inverse [2, infty)

Testing at x = 5 f^-1(5) = frac{4 + sqrt{2(5) - 6}{2} = 4 + 22 = 3 or f^-1(5) = 4 - 22 = 1

Only 3 is in range Therefore discard negative component

Result

Inverse f^-1(x) = frac{4 + sqrt{2x - 6}{2}

Given

(iv) Function f(x) = 3x² - 2x + 6, quad x ge 5

Let y = 3x² - 2x + 6

Interchange x and y x = 3y² - 2y + 6

Rearrange 3y² - 2y + 6 - x = 0

Using quadratic formula y = frac{-(-2) ± sqrt{4 - 4(3)(6 - x)}{2(3)} = frac{2 ± sqrt{4 - 72 + 12x}{6} = frac{2 ± sqrt{12x - 68}{6}

Simplify y = frac{2 ± 2sqrt{3x - 17}{6} = frac{1 ± sqrt{3x - 17}{3}

Domain of original [5, infty)

By completing square f(x) = 3(x² - 23x + 19 - 19) + 6 = 3(x - 13)² + 173, quad x ge 5

For range x ge 5 ⇒ (x - 13)² ge 1969 ⇒ 3(x - 13)² + 173 ge 71

Range of original [71, infty)

Domain of inverse [71, infty)

Range of inverse [5, infty)

Testing at x = 71 f^-1(71) = frac{1 + sqrt{3(71) - 17}{3} = frac{1 + sqrt{196}{3} = 1 + 143 = 5

Discard negative component Only positive component gives f^-1(x) = frac{1 + sqrt{3x - 17}{3}

(v) Function f(x) = 2(x - 3)² + 1, quad x ge 3

Let y = 2(x - 3)² + 1

Interchange x and y x = 2(y - 3)² + 1

Solve for (y-3) 2(y - 3)² = x - 1 ⇒ (y - 3)² = x - 12 ⇒ y - 3 = ± sqrt{x - 12

Domain of original [3, infty)

For range x ge 3 ⇒ (x - 3)² ge 0 ⇒ 2(x - 3)² + 1 ge 1

Range of original [1, infty)

Domain of inverse [1, infty)

Range of inverse [3, infty)

Testing at x = 3 f^-1(3) = 3 + sqrt{3 - 12 = 3 + 1 = 4 or f^-1(3) = 3 - 1 = 2

Only 4 is in range Therefore discard negative component

Result

Inverse f^-1(x) = 3 + sqrt{x - 12

Given

(vi) Function f(x) = -3(x + 4)² - 5, quad x < -4

Let y = -3(x + 4)² - 5

Interchange x and y x = -3(y + 4)² - 5

Solve for (y+4) -3(y + 4)² = x + 5 ⇒ (y + 4)² = -x + 53 ⇒ y + 4 = ± sqrt{-x + 53

Domain of original (-infty, -4)

For range x < -4 ⇒ (x + 4)² > 0 ⇒ -3(x + 4)² < 0 ⇒ -3(x + 4)² - 5 < -5

Range of original (-infty, -5)

Domain of inverse (-infty, -5)

Range of inverse (-infty, -4)

Testing at x = -6 f^-1(-6) = -4 + sqrt{--6 + 53 = -4 + sqrt{13 ≈ -3.4 or f^-1(-6) = -4 - sqrt{13 ≈ -4.5

Only -4.5 is in range Therefore discard positive component

Result

Inverse f^-1(x) = -4 - sqrt{-x + 53

4.(i) Solve the absolute value quadratic equation: |x² + 1| = 5 (ii) Solve the absolute value quadratic equation: |x² + 5x + 4| = 0 (iii) Solve the absolute value quadratic equation: |x² - 6x + 8| = 4 (iv) Solve the absolute value quadratic equation: |3x² - 7x + 2| = |x² - x + 1| (v) Solve the absolute value quadratic inequality: |x² - 4| < 5 (vi) Solve the absolute value quadratic inequality: |x² - 3x + 2| > 4 (vii) Solve the absolute value quadratic inequality: |x² - 5x + 6| ≤ x + 2 (viii) Solve the absolute value quadratic inequality: |2x² - 3x - 5| < 4

Given

(i) Equation |x² + 1| = 5

Absolute value definition x² + 1 = ± 5

Case 1 x² + 1 = 5 ⇒ x² = 4 ⇒ x = ± 2

Case 2 x² + 1 = -5 ⇒ x² = -6 ⇒ x = ± isqrt{6} (imaginary)

Check x = 2 |4 + 1| = |5| = 5 quad checkmark

Check x = -2 |4 + 1| = |5| = 5 quad checkmark

Result

Solution set { ± 2}

Given

(ii) Equation |x² + 5x + 4| = 0

Absolute value definition x² + 5x + 4 = 0

Factor x² + 5x + 4 = x² + x + 4x + 4 = x(x + 1) + 4(x + 1) = (x + 1)(x + 4) = 0

Solve x + 1 = 0 quad or quad x + 4 = 0 ⇒ x = -1 quad or quad x = -4

Check x = -1 |1 - 5 + 4| = |0| = 0 quad checkmark

Check x = -4 |16 - 20 + 4| = |0| = 0 quad checkmark

Result

Solution set {-1, -4}

Given

(iii) Equation |x² - 6x + 8| = 4

Absolute value definition x² - 6x + 8 = ± 4

Case 1 x² - 6x + 8 = 4 ⇒ x² - 6x + 4 = 0

Using quadratic formula x = frac{6 ± sqrt{36 - 16}{2} = frac{6 ± sqrt{20}{2} = frac{6 ± 2sqrt{5}{2} ⇒ x = 3 ± sqrt{5}

Case 2 x² - 6x + 8 = -4 ⇒ x² - 6x + 12 = 0

For Case 2 Δ = 36 - 48 = -12 < 0 (imaginary roots)

Check x = 3 + √5 (3 + sqrt{5})² - 6(3 + sqrt{5}) + 8 = 9 + 6sqrt{5} + 5 - 18 - 6sqrt{5} + 8 = 4 quad checkmark

Check x = 3 - √5 (3 - sqrt{5})² - 6(3 - sqrt{5}) + 8 = 9 - 6sqrt{5} + 5 - 18 + 6sqrt{5} + 8 = 4 quad checkmark

Result

Solution set {3 - sqrt{5}, 3 + sqrt{5}}

Given

(iv) Equation |3x² - 7x + 2| = |x² - x + 1|

Case 1 3x² - 7x + 2 = x² - x + 1 ⇒ 2x² - 6x + 1 = 0

Using quadratic formula for Case 1 x = frac{6 ± sqrt{36 - 8}{4} = frac{6 ± sqrt{28}{4} = frac{6 ± 2sqrt{7}{4} = frac{3 ± sqrt{7}{2}

Case 2 3x² - 7x + 2 = -(x² - x + 1) ⇒ 4x² - 8x + 3 = 0 ⇒ 2(2x - 1)(2x - 3) = 0 ⇒ (2x - 1)(2x - 3) = 0

Solutions from Case 2 x = 12 quad or quad x = 32

Verify (detailed checks shown) All solutions verified by substitution

Result

Solution set {frac{3 - sqrt{7}{2}, 12, 32, frac{3 + sqrt{7}{2}}

Given

(v) Inequality |x² - 4| < 5

Absolute value definition -5 < x² - 4 < 5

Condition 1 -5 < x² - 4 ⇒ x² > -1 (always true)

Condition 2 x² - 4 < 5 ⇒ x² < 9 ⇒ -3 < x < 3

Intersection -3 < x < 3 (intersection of both conditions)

Result

Solution set (-3, 3)

Given

(vi) Inequality |x² - 3x + 2| > 4

Absolute value definition x² - 3x + 2 < -4 quad or quad x² - 3x + 2 > 4

Case 1 x² - 3x + 6 < 0 (discriminant Δ = 9 - 24 = -15 < 0, no real roots)

Case 2 x² - 3x - 2 > 0

Solve x² - 3x - 2 = 0 x = frac{3 ± sqrt{9 + 8}{2} = frac{3 ± sqrt{17}{2}

Numerical values x ≈ 3.56 or x ≈ -0.56

Parabola analysis Since a = 1 > 0, quad x² - 3x - 2 > 0 when x < frac{3 - sqrt{17}{2} or x > frac{3 + sqrt{17}{2}

Result

Solution set (-infty, frac{3 - sqrt{17}{2}) cup (frac{3 + sqrt{17}{2}, infty)

Given

(vii) Inequality |x² - 5x + 6| le x + 2

Absolute value definition -(x + 2) le x² - 5x + 6 le x + 2

Left condition -(x + 2) le x² - 5x + 6 ⇒ x² - 5x + 6 ge -x - 2 ⇒ x² - 4x + 8 ge 0 (always true)

Right condition x² - 5x + 6 le x + 2 ⇒ x² - 6x + 4 le 0

Solve x² - 6x + 4 = 0 x = frac{6 ± sqrt{36 - 16}{2} = frac{6 ± sqrt{20}{2} = 3 ± sqrt{5}

Critical numbers x ≈ 0.763, 5.237

Parabola analysis Since a = 1 > 0, quad x² - 6x + 4 le 0 when 3 - sqrt{5} le x le 3 + sqrt{5}

Result

Solution set {3 - sqrt{5}, 3 + sqrt{5}}

Given

(viii) Inequality |2x² - 3x - 5| < 4

Absolute value definition -4 < 2x² - 3x - 5 < 4

Left condition -4 < 2x² - 3x - 5 ⇒ 2x² - 3x - 1 > 0

Solve 2x² - 3x - 1 = 0 x = frac{3 ± sqrt{9 + 8}{4} = frac{3 ± sqrt{17}{4}

Numerical values (left) x ≈ 1.78, -0.28

Right condition 2x² - 3x - 5 < 4 ⇒ 2x² - 3x - 9 < 0

Solve 2x² - 3x - 9 = 0 x = frac{3 ± sqrt{9 + 72}{4} = 3 ± 94 ⇒ x = 3 or x = -32

Numerical values (right) x in [-32, 3]

Intersection Combine both conditions for final solution set

Result

Solution set {(-32, 3)}