Zeros Of Cubic Function

Prakhar Birla - Grade 12 C

Zeros of cubic function - Type I

Mathematics Portfolio Type I – By Prakhar Birla Grade 12 C Page 1

Q1. Consider the cubic equation: f(x) = 2x3 + 6x2 – 4.5x – 13.5 … A1. The graph: y

f(x)=2x^3 + 6x^2 - 4.5x - 13.5

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x -4

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Window values: x Min: -4.3556517 x Max: 2.98128718 y Min: -17.61584636 y Max: 7.191107

The roots of the equation are: • x = 1.5 • x = -1.5 • x = -3.0 Proof using the remainder theorem: • f(x) = 2x3 + 6x2 – 4.5x –13.5 f(1.5) = 2(1.5)3 + 6(1.5)2 – 4.5(1.5) – 13.5 f(1.5) = 0 • f(x) = 2x3 + 6x2 – 4.5x -13.5 f(-1.5) = 2(-1.5)3 + 6(-1.5)2 – 4.5(-1.5) –13.5 f(-1.5) = 0 • f(x) = 2x3 + 6x2 – 4.5x -13.5 f(-3.0) = 2(-3.0)3 + 6(-3.0)2 – 4.5(-3.0) –13.5 f(-3.0) = 0 Equations of tangent lines (average of two of roots): • Roots 1.5 and -1.5 Average = [1.5 + (-1.5)]/2 = 0 f(0) = 2(0)3 + 6(0)2 – 4.5(0) –13.5 f(0) = -13.5 f `(x) = 6x2 + 12x – 4.5 Slope at (0, -13.5) will be: f `(0) = 6(0)2 + 12(0) – 4.5 = -4.5 Equation of tangent: y – y1 = m(x – x1), where x1= 0; y1= -13.5; m= -4.5 y = -4.5x – 13.5

Mathematics Portfolio Type I – By Prakhar Birla Grade 12 C Page 2

y

f(x)=2x^3 + 6x^2 - 4.5x - 13.5 y=-4.5x-13.5

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x -4

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-1

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-5

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•

Roots 1.5 and -3.0 Average = [1.5 + (-3.0)]/2 = -0.75 f(-1.5) = 2(-1.5)3 + 6(-1.5)2 – 4.5(-1.5) –13.5 f(-1.5) = -7.59 ≈ -7.6 Slope at (-0.75, -7.6) will be: f `(-0.75) = 6(-0.75)2 + 12(-0.75) – 4.5 = -10.125 Equation of tangent: y – y1 = m(x – x1), where x1= -1.5; y1= -7.6; m= -10.125 y = -10.125x – 15.2 y

f(x)=2x^3 + 6x^2 - 4.5x - 13.5 y=-4.5x-13.5

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x -4

-3

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Roots -1.5 and -3.0 Average = [(-1.5) + (-3.0)]/2 = -2.25 f(-2.25) = 2(-2.25) 3 + 6(-2.25) 2 – 4.5(-2.25) –13.5 f(-2.25) = -4.205 ≈ -4.2 Slope at (-2.25, 4.2) will be: f `(-2.25) = 6(-2.25)2 + 12(-2.25) – 4.5 = -1.125 Equation of tangent: y – y1 = m(x – x1), where x1= -2.25; y1= -4.2; m= -1.125 y = -1.125x + 1.7

Mathematics Portfolio Type I – By Prakhar Birla Grade 12 C Page 3

y

f(x)=2x^3 + 6x^2 - 4.5x - 13.5 y=-1.125x+1.6875

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x -4

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-1

1

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Intersection of equations of tangent with curve: • y = -4.5x – 13.5 0 = -4.5x – 13.5 x = -3.0 Hence intersection will be (-3.0, 0) • y = -10.125x – 15.2 0 = -10.125x – 15.2 x = -1.501 ≈ -1.5 Hence intersection will be (-1.5, 0) • y = -1.125x + 1.7 0 = -1.125x + 1.7 x = 1.511 ≈ 1.5 Hence intersection will be (1.5, 0) Observation: The equation of tangent at the average of any two roots intersects the y–axis at the other root i.e. the root that is not included in the average. This observation holds for all three examined cases. Conjecture: The equation of tangent at the average of any two roots will intersects the y–axis at the third root i.e. the root that is not included in the average.

Mathematics Portfolio Type I – By Prakhar Birla Grade 12 C Page 4

Q2. Test your conjecture in similar other cubic functions. A2. Test Equation: g(x) = x3 + 10x2 – 2x – 108 y

f(x)=x^3 + 10x^2 - 2x - 108

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x -25

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-100

The roots of the equation are: • x = -9 • x = -4 • x=3 g `(x) = 3x2 + 20x – 2 Equations of tangent lines (average of two of roots): • Roots -9 and -4 Average = [(-9) + (-4)]/2 = -6.5 g(-6.5) = (-6.5)3 + 10(-6.5)2 – 2(-6.5) – 108 g(-6.5) = 59.375 Slope at (-6.5, 59.375) will be: g `(-6.5) = 3(-6.5)2 + 20(-6.5) – 2 = -6.25 Intersection of equations of tangent with curve: y – y1 = m(x – x1), where x1= -6.5; y1= 59.375; m= -6.25 At y=0, x=3

Mathematics Portfolio Type I – By Prakhar Birla Grade 12 C Page 5

y

f(x)=x^3 + 10x^2 - 2x - 108 y=-5.25x+18.75

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x -25

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•

Roots -9 and 3 Average = [(-9) + 3]/2 = -3 g(-3) = (-3)3 + 10(-3)2 – 2(-3) – 108 g(-3) = -36 Slope at (-3, -36) will be: g `(-3) = 3(-3)2 + 20(-3) – 2 = -36 Intersection of equations of tangent with curve: y – y1 = m(x – x1), where x1= -3; y1= -36; m= -36 At y=0, x=-4 y

f(x)=x^3 + 10x^2 - 2x - 108 y=-35x-144

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Roots -4 and 3 Average = [(-9) + 3]/2 = -0.5 g(-0.5) = (-0.5)3 + 10(-0.5)2 – 2(-0.5) – 108 g(-0.5) = -104 Slope at (-0.5, -36) will be: g `(-0.5) = 3(-0.5)2 + 20(-0.5) – 2 = -12.3 Intersection of equations of tangent with curve: y – y1 = m(x – x1), where x1= -0.5; y1= -104; m= Mathematics Portfolio Type I – By Prakhar Birla Grade 12 C Page 6

-12.3 At y=0, x=-9 y

f(x)=x^3 + 10x^2 - 2x - 108 y=-11.25x-110.25

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With this we can conclude that our conjecture stands for other cubic equations as well. Q3. Prove your conjecture. A3. Let f(x) = k(x-a)(x-b)(x-c)  ab Taking the average of two roots i.e. root a and root b at x    2  Then y will be   ab   ab   ab   yk   a   b   c    2   2     2 Simplifying the equation:   a  b  2a   a  b  2b   a  b  2c   k     2 2 2          b  a   b  a   a  b  2c  

k



8    a  b

 k  

2



 a  b  2c   8

 

Mathematics Portfolio Type I – By Prakhar Birla Grade 12 C Page 7

As y = k (x  a )(x  b)(x  c) ln y  ln k  ln( x  a )  ln( x  b)  ln( x  c) 1 1 1 1 f '( x)  0    y xa xb xc So, f ( x)  k  ( x  b)( x  c )  ( x  a )( x  c)  ( x  a)( x  b)    ab    ab    ab     ab     ab     ab    ab   p     a     b      a     c      b     c    2     2     2     2     2     2       2    a  r  2a   a  b  2b   a  b  2a   a  b  2c   a  b  2b   a  b  2c   k          2 2 2 2 2 2             

f '

    a  b  2    b  a   a  b  2c     a  b   a  b  2c   k       4 4 4            a  b  2    b  a   a  b  2c     b  a   a  b  2c   k       4 4 4       

     

  a  b 2 

 k  

4

 

So the equation is:  k (a  b) 2 (a  b  2c)   kx(a  b) 2 a( a  b)(a  b) 2 y     8 4 8   kx (a  b) 2 k (a  b)(a  b) 2 a(a  b) 2 (a  b  2c)   4 8 8 2 2 2 kx (a  b) k (a  b)(a  b)  k (a  b) (a  b  2c) y  4 8 2 kx(a  b)  k (a  b)  k (a  b  2c)   ( a  b) 2   4 8  y

As y is zero when the tangent cuts the x-axis

 k (a  b)  k (a  b  2c)   2  2 px  ka  kb  ka  kb  2kc 2kx  2kc Or, x  c As x = c, which is the third i.e. the root not included in the average, we can conclude that our conjecture is true for all cubic equations. Restating it “The equation of tangent at the average of any two roots will intersects the y–axis at the third root i.e. the root that is not included in the average.” kx  

Mathematics Portfolio Type I – By Prakhar Birla Grade 12 C Page 8

Q4. Investigate the above properties with cubic equations which have: (a) one root (b) two roots (c) one real and two complex roots. A4. (a) Let us consider the equation g(x) = 6x3 + 7x2 + 8x + 9

Mathematics Portfolio Type I – By Prakhar Birla Grade 12 C Page 9

y

f(x)=(6) x^3 + (7) x^2 + (8) x + (9)

8 6 4 2

x -8

-6

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-2 -4 -6 -8

The root of the equation is:  x = -1.1456 Proof using the remainder theorem:  f(x) = 6x3 + 7x2 + 8x + 9 f(-1.1456) = 6(-1.1456)3 + 7(-1.1456)2 + 8(-1.1456) + 9 f(-1.1456) = 0 Equation of the tangent line (average of two of roots):  Roots -1.1456 and -1.1456 Average = [(-1.1456) + (-1.1456)]/2 = -1.1456 f(-1.1456) = 6(-1.1456)3 + 7(-1.1456)2 + 8(-1.1456) + 9 f(-1.1456) = 0 f `(x) = 18x2 + 14x + 8 Slope at (-1.1456, 0) will be: f `(-1.1456) = 18(-1.1456)2 + 14(-1.1456) + 8 = 15.6 Equation of tangent: y – y1 = m(x – x1), where x1= -1.1456; y1= 0; m= 15.6 At y = 0, x = -1.1456 i.e. the third root (as all three roots are equal)

Mathematics Portfolio Type I – By Prakhar Birla Grade 12 C Page 10

y

f(x)=(6) x^3 + (7) x^2 + (8) x + (9) y=15.5848x+17.855

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-5 -10 -15 -20

This shows that the conjecture is applicable to a cubic equation with one root, although not required. (b) Let us consider the equation g(x) = x3 + 4 x2 + 5x + 2 y

f(x)=(1) x^3 + (4) x^2 + (5) x + (2)

8 6 4 2

x -8

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-2 -4 -6 -8

The roots of the equation are: • x = -2 • x = -1 • x = -1 Proof using the remainder theorem: • f(x) = x3 +4 x2 + 5x + 2 f(-1) = (-1)3 + 4(-1)2 + 5(-1) + 2 f(-1) = 0

Mathematics Portfolio Type I – By Prakhar Birla Grade 12 C Page 11

•

f(x) = x3 +4 x2 + 5x + 2 f(-2) = (-2)3 + 4(-2)2 + 5(-2) + 2 f(-2) = 0

Equation of the tangent line (average of two of roots): • Roots -2 and -1 Average = [(-1) + (-2)]/2 = -1.5 f(-1.5) = (-1.5)3 + 4(-1.5)2 + 5(-1.5) + 2 f(-1.5) = 0.125 f `(x) = 3x2 + 7x + 5 Slope at (-1.5, 0.125) will be: f `(-1.1456) = 3(-1.1456)2 + 7(-1.1456) + 5 = -0.25 Equation of tangent: y – y1 = m(x – x1), where x1= -1.5; y1= 0.125; m= -0.25 At y = 0, x = -1 i.e. the third root y

f(x)=(1) x^3 + (4) x^2 + (5) x + (2) y=0x+0

8 6 4 2

x -8

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-2 -4 -6 -8

This again shows that the conjecture is applicable to a cubic equation with two roots. (b) Let us consider the equation g(x) = x3 + x2 + x + 1 y

f(x)=(6) x^3 + (9) x^2 + (3) x + (68)

8 6 4 2

x -8

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-2 -4 -6 -8

The roots of the equation are: • x = -1 • x = 1i Mathematics Portfolio Type I – By Prakhar Birla Grade 12 C Page 12

•

x = -1i

Proof using the remainder theorem: • f(x) = x3 + x2 + x + 1 f(-1) = (-1)3 + (-1)2 + (-1) + 1 f(-1) = 0 • f(x) = x3 + x2 + x + 1 f(1i) = (1i)3 + (1i)2 + (1i) + 1 f(1i) = 0 • f(x) = x3 + x2 + x + 1 f(-1i) = (-1i)3 + (-1i)2 + (-1i) + 1 f(-1i) = 0 Equation of the tangent line (average of two of roots): • Roots -1 and 1i Average = [(-1) + (-1i)]/2 = -0.5 – 0.5i f(-0.5 – 0.5i) = (-0.5 – 0.5i)3 + (-0.5 – 0.5i)2 + (-0.5 – 0.5i) + 1 f(-0.5 – 0.5i) = 0.75 + 0.25i f `(x) = 3x2 + 2x + 1 Slope at (-0.5 – 0.5i, 0.75 + 0.25i) will be: f `(-0.5 – 0.5i) = 3(-0.5 – 0.5i)2 + 1(-0.5 – 0.5i) + 1 = -0.5i Equation of tangent: y – y1 = m(x – x1), where x1= -0.5 – 0.5i; y1= 0.75 + 0.25i; m= -0.5i At y = 0, x = -i i.e. the third root Hence this cubic equation also complies with the conjecture stated.

“The equation of tangent at the average of any two roots will intersects the y–axis at the third root i.e. the root that is not included in the average.” Mathematics Portfolio Type I – By Prakhar Birla Grade 12 C Page 13

Bibliography 1. Graphing Software – Graph v1.4 2. Document Editor – Microsoft Word 2003 3. Advanced Equation Editor – Design Science MathType 5.2c

Mathematics Portfolio Type I – By Prakhar Birla Grade 12 C Page 14