Zeros Of Cubic Functions

MATHEMATICS IB HIGHER PORTFOLIO

ZEROS OF CUBIC FUNCTIONS FRIDAY, MAY 20, 2005

Peter Hamilton

Zeros of cubic functions HL Type I Investigation Consider the cubic function: f x=2 x 36 x 2−4.5 x−13.5

Graphed on a Ti-86 calculator:

The roots of this equation: x=−3

x=−1.5

x=1.5

In order to find the tangent at any given point on the graph, the slope at that point must be found. To find the slope at any given point, the first differential of the function can be used: f x=2 x 36 x 2−4.5 x−13.5 f 1  x=6 x 212 x−4.5 To find the tangent line of the average of two of the three roots, the value must be plugged into the first differential equation in order to find its slope. Taking roots -3 and 1.5: f 1 x=6 x 212 x−4.5 Average of -3 and 1.5 −31.5 =−0.75 2 f 1 −0.75=6−0.75212−0.75−4.5=−10.125 f 1 −0.75=−10.125 Taking roots -3 and -1.5:

f 1 x=6 x 212 x−4.5 Average of -3 and -1.5 −3−1.5 =−2.25 2 f 1 −2.25=6−2.25212−2.25−4.5=−1.125 f 1 −2.25=−1.125 Taking roots 1.5 and -1.5: f 1 x=6 x 212 x−4.5 Average of 1.5 and -1.5 1.5−1.5 =0 2 f 1 0=602120−4.5=−4.5 f 1 0=−4.5 A tangent line can be defined by: y=m xb By using the values obtained for the slope of these tangent lines, the next values necessary are the respective y values in each case. These can be determined by plugging the averages into the original equation. Roots -3 and 1.5: 3

2

f x=2 x 6 x −4.5 x−13.5 f −0.75=2−0.7536−0.752−4.5−0.75−13.5=−7.59375 f 0.75=−7.59375 Roots -3 and -1.5: f x=2 x 36 x 2−4.5 x−13.5 f −2.25=2−2.2536−2.252−4.5−2.25−13.5=4.21875 f 2.25=4.21875 Roots 1.5 and -1.5 f x =2 x 36 x 2−4.5 x−13.5 f 0=203602−4.50−13.5 f 0=−13.5

Using the equation: y=mx+b and inserting the corresponding values which have been calculated, the equations of the tangent lines can be found. Roots -3 and 1.5: y=mxb y=−7.59375 m=−10.125 x=−0.75 −7.59375=−10.125 −0.75b b=−15.1875 y=−10.125 x−15.1875 Roots -3 and -1.5: y=mxb y=4.21875 m=−1.125 x=−2.25 4.21875=−1.125−2.25b b=1.6875 y=−1.125 x1.6875 Roots 1.5 and -1.5 y=mxb y=−13.5 m=−4.5 x=0 −13.5=−4.50b b=−13.5 y=−4.5 x−13.5 Now, by examining the x intercepts of each equation, a statement concerning the nature of these intercepts can be made: Roots -3 and 1.5: y=−10.125 x−15.1875 Let y=0 0=−10.125 x−15.1875 15.1875=−10.125 x x=−1.5 Roots -3 and -1.5:

y=−1.125 x1.6875 Let y=0 0=−1.125 x1.6875 −1.6875=−1.125 x x=1.5 Roots 1.5 and -1.5 y=−4.5 x−13.5 Let y=0 0=−4.5 x−13.5 13.5=−4.5 x x=−3 The following graph demonstrates the route of each tangent line of the function:

In each given scenario, the line tangent to the average of two roots of a cubic equation also passes through the third root. To further demonstrate this conjecture, similar cubic functions can be taken and the same result may be found. example 1 f x=x 3 −2 x 2−5 x6 Roots at: x=−2 x=1 x=3

Roots -2 and 3 −23 =0.5 2 3 2 f 0.5=0.5 −20.5 −50.56=3.125 f 1 x=3 x 2−4 x−5 f 1 0.5=30.52−40.5−5=−6.25 y=mxb 3.125=−6.250.5b b=6.25 y=−6.25 x6.25 Let y=0 0=−6.25 x6.25 x=1 Roots 1 and 3 13 Average value: =2 2 3 2 f 2=2 −22 −526=−4 f 1 x=3 x 2−4 x−5 1 2 f 2=32 −42−5=−1 y=mxb −4=−12b b=−2 y=−x±2 Let y=0 0=−x−2 x=−2 Roots -2 and 1 −21 Average value: =−0.5 2 f −0.5=−0.53−2−0.52−5−0.56=7.875 1 2 f x =3 x −4 x−5 f 1 −0.5=3−0.52−4−0.5−5=−2.25 y=mxb 7.875=−2.25−0.5b b=6.75 y=−2.25 x6.75 x=3 Let y=0 0=−6.25 x6.25 x=1 Average value:

Demonstrated graphically:

Example 2 3 2 f  x=x −3 x −4 x12 Roots at: x=−2 x=2 x=3

Roots -2 and 2 −22 =0 2 3 2 f 0=0 −30 −4012=12 1 2 f x=3 x −6 x−4 f 1 0=302−60−4=−4 y=mxb 12=−40b b=12 y=−4 x12 Let y=0 0=−4 x12 x=−3 Average value:

Roots -2 and 3 −23 =0.5 2 3 2 f 0.5=0.5 −30.5 −40.512=9.375 f 1 x=3 x 2−6 x−4 f 1 0.5=30.52−60.5−4=−6.25 y=mxb 9.375=−6.250.5b b=12.5 y=−6.25 x12.5 Let y=0 0=−6.25 x12.5 x=−2 Roots 2 and 3 23 Average value: =2.5 2 3 2 f 2.5=2.5 −32.5 −42.512=−1.125 f 1 x=3 x 2−6 x−4 1 2 f 2.5=32.5 −62.5−4=−0.25 y=mxb −1.125=−0.252.5b b=−0.5 y=−0.25 x−0.5 Let y=0 0=−0.25 x−0.5 x=−2 Demonstrated graphically: Average value:

These examples are consistent with the idea that the tangent to the average of any two roots of a cubic passes through the third root. In order to prove this as a property for all roots, a general case must be taken and proven true for all instances.

A function with roots case.

 ,  , and  can be used to provide a general

f x=x−x−x− =x 2− x− x x− = x 3− x 2− x 2  x− x 2  x  x−   f x=x 3− x 2    x−   As multiplication is commutative, it can be said that f x=x−x−x− is equivalent to f x= x−x−x− and therefore only one case is necessary in proving that the tangent to the average of any two roots will pass through the third root.  into the first differential we can find 2 the slope of the average of two roots of a cubic.

By substituting the value

f 1 x=3 x 2−2 x      2  f 1 =3  −2     2 2 2 3 = 2−    4 3 4 4 = 22  2 − 2   2      4 4 4 2 2 2 2 3  6  3  −4  −4  −4  −4  −8  4  4  4   = 4 2 2 2 − 2  −  −− 1  f  = = 2 4 4 2 −− 1  f  = 2 4 Additionally, the y value of the average of two roots in necessary in order to find the tangent. This is found by plugging the value  into the original function. 2

f x=x 3− x 2    x−     3  2  f =  −      −   2 2 2 2  3 33 2 3  232 2   2  = − 2 4 2 2 2 2   2       2     − 2 2 3 3 2 2 3 2   3  3     2   2  2 2  22  = −  8 4 2 3 2 2 3 2 2  −    − 2  −4   2   f = 2 8 This value, along with that obtained with the first differential can be used to find the tangent to the average of two roots. By substituting the corresponding slope and y-value, the tangent can be solved for. y=mxb −3 2  2−32 2 −4   2 2  Let y= 8 2 −− m= 4  x= 2 3 2 −   2−32 2 −4   2 2  −−2  = b 8 4 2 −32  2−32 2 −4   2 2 3− 2−2 3  b= 8 2 2 2  −4   2   b= 8 2 2 2 −− 2  −4   2   y= x 4 8 With the tangent line completed, the test to the conjecture can be shown by substituting the root  , 0 into equation. If the equation holds true, then the conjecture has been proven for all cubics.

−−2 2 2 −4   2 2  y= x 4 8 Let y=0 x= −−2 2 2 −4   2 2  0=  4 8 2 2 2  −2    2  −4   2 2  0=−2  8 8 2 2 2 2 −2  4   −2  2  −4   2   0= 8 0 0= =0 8 Therefore this conjecture has been proved true for all cubic functions which can be expressed as x−x−x− . There exist certain cubic functions which contain roots unlike the previous examples. Some cubics only have one root, others have only two roots, and others have one real root and two imaginary ones. The possibility of a cubic having one root is illustrated with the following example: f x=x 33 x 23 x1 Root x=1 This equation has 3 identical roots Therefore, the average of any two roots will result in the same calculation 11 Average: =0 2 f 1=13=1 f 1 x=2 x 2 f 1 1=212=2 y=mxb Let y=1 x=1 m=2 1=12b b=−1 y=x−1 To test the conjecture, it is necessary to test the final root in the tangent equation Let x=1 y=0 0=1−1=0 Proven True

A cubic with only two roots follows the same pattern. In reality it has 3 roots, with 2 of them being identical roots. For example:

Roots

x=3

f  x=x 3 x 2 −8 x−12 x=−2 with a double root at x=−2

Roots 3 and -2 3±2 Average =0.5 2 f 0.5=0.530.52−80.5−12=−15.625 f 1  x=3 x 22 x−8 1 2 f 0.5=30.5 20.5−8=−6.25 y=mxb Let y=−15.625 x=0.5 m =−6.25 −15.625=−6.25.5b b =−12.5 Therefore the tangent equation is: y=−6.25 x−12.5 By substiting the x and y values of the third root of the cubic, the conjecture can be tested Let y=0 x=−2 0=−6.25−2−12.5=0 Proven True Roots -2 and -2 −2±2 Average =−2 2 3 2 f −2=−2 −2 −8−2−12=0 f 1  x=3 x 22 x−8 f 1 −2=3−222−2−8=0 y=mxb Let y=0 x=−2 m =0 0=0 −2b b =0 Therefore the tangent equation is: y=0 By substiting the x and y values of the third root of the cubic, the conjecture can be tested Let y=0 x=3 0=0−2−0=0 Proven True

Two Roots Shown Graphically

Thus, it has been shown that for cubics with either a double or triple root follows the conjecture stated above. The final area of investigation is beyond the domain of real numbers. A cubic with one real and two complex roots will also adhere to the conjecture, as is shown in the following example:

f  x= x 3− x 2 x−1 Roots x=i x=−i x=1 Roots i and 1 1i  Average: 2 1i  1i  3 1i  2 1i  f  =  −  −1 2 2 2 2 13 i −3−i  12 i −1 1i  2 = −  − 8 4 2 2 13 i −3−i −2−4 i 244 i −8 = 8 1i  −62 i  f  = 2 8 1 2 f  x=3 x −2 x1 1i  1i  2 1i  f 1 =3  −2 1 2 2 2 1i  2 i  1i  4 f 1 =3 −4  2 4 4 4 2 i  1 1i  f  = 2 4 Using the value of slope and the x and y values of the average of the two roots, the tangent line can be solved for:

y=mxb −62 i  1i  2 i  Let y= x= m= 8 2 4 −62 i  2 i  1i  = b 8 4 2 −4 b = 8 2 i  −4 y= x 4 8 If the third root is subsituted the equation should still be true, proving the conjecture Let y=0 x=−i 2 i  −4 0= −i  4 8 4 −4 0=  8 8 4−4 0= =0 8 Proven True Roots −i and 1 1−i  Average: 2 3 2 1−i  1−i  1−i  1−i  f  =  −  −1 2 2 2 2 1−3 i −3i  1−2 i −1 1−i  2 = −  − 8 4 2 2 1−3 i −3i −24 i 24−4 i −8 = 8 1−i  −6−2 i  f  = 2 8 1 2 f  x=3 x −2 x1 2 1−i  1−i  1 1−i  f  =3  −2 1 2 2 2 1−i  −2 i  1−i  4 f 1 =3 −4  2 4 4 4 −6 i −44 i 4 −i 1 1−i  f  = = 2 2 2 Using the value of slope and the x and y values of the average of the two roots, the tangent line can be solved for.

y=mxb −3−i  1−i  −i  Let y= x= m= 4 2 2 −3−i  −i  1−i  = b 4 2 2 −3−i i 1 −1 b = = 4 2 −i  1 y= x− 2 2 If the third root is subsituted the equation should still be true, proving the conjecture Let y=0 x=i −i  1 0= i − 2 2 1 1 0= − 2 2 1−1 0= =0 2 Proven True Roots −i and i i −i  Average: =0 2 f 0=03−0 20−1=−1 f 1  x=3 x 2 −2 x1 f 1 0=302 −201 f 1 0=30−401=1 y=mxb Let

y=−1

x=0

m =1

−1=10b b =−1 y= x−1 If the third root is subsituted, the equation should still be true, proving the conjecture Let y=0 x=1 0=1−1=0 Proven True Graph and real tangents shown graphically

In conclusion, it is evident that this conjuncture has proved true. For a cubic in form f  x= x− x− x− the tangent at the average of any two roots will pass through the third root. It has been shown true for various cubics, been demonstrated true for a general case, and verified with examples including one root, two roots, and one real and two complex roots.