John McCarthy
Investigation of Cubic Functions
Year 12 Maths Studies: Graphs of Cubic Functions Investigation
John McCarthy
Introduction:
A cubic polynomial is any function of the form 𝑦 = 𝑎𝑥 3 + 𝑏𝑥 2 +
𝑐𝑥 + 𝑑, 𝑎 ≠ 0 and exhibits both a concave and convex graph region.
The following investigation aims to investigate the relationship
between the two stationary points of a cubic polynomial and the
point of inflection of that function. It will investigate situations
where the leading coefficient of the function is equal, or not equal
to 1, and situations where the expressions found do not apply.
Finding this relationship will result in a simpler way of calculating
the point of inflection of a cubic polynomial without having to find
the second derivative.
An example of a Cubic Polynomial
Investigating Cubic Functions where the first coefficient is equal to 𝟏:
The goal of this section is to investigate the nature of the two stationary points and the single point
of inflection for cubic functions where the leading coefficient is equal to 1. Letting 𝐴, 𝐵 and 𝐺 equal
the two stationary points and the point of inflection respectively, we can find these values for a set
of sample functions.
For 𝒚 = 𝒙𝟑 − 𝟑𝒙𝟐 − 𝟗𝒙 + 𝟕:
𝑑𝑦
= 3𝑥 2 − 6𝑥 − 9
𝑑𝑥
𝑑2 𝑦
= 6𝑥 − 6
𝑑𝑥 2
To find the two stationary points of this function we must find when
𝑑𝑦
𝑑𝑥
= 0.
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John McCarthy
Investigation of Cubic Functions
3𝑥 2 − 6𝑥 − 9 = 0
𝑥=
6 ± √36 − 4 × 3 × −9
= 3 𝑜𝑟 − 1
6
Hence 𝐴 = −1, 𝐵 = 3.
𝑑2 𝑦
To find the point of inflection we must find when 𝑑𝑥 2 = 0.
6𝑥 − 6 = 0
𝑥=1
Substituting these 𝑥 values back into the original equation, we find the coordinates:
𝐴 = (−1, 12), 𝐵 = (3, −20), 𝐺 = (1, −4).
For 𝒚 = 𝒙𝟑 − 𝟏𝟐𝒙𝟐 + 𝟐𝟏𝒙 − 𝟏𝟒:
Calculations are found in the Appendix.
𝐴 = (1, −4), 𝐵 = (7, −122), 𝐺 = (4, −58)
For 𝒚 = 𝒙𝟑 + 𝟗𝒙𝟐 − 𝟏𝟐:
Calculations are found in the Appendix.
𝐴 = (−6,96), 𝐵 = (0, −12), 𝐺 = (−3,42)
For 𝒚 = 𝒙𝟑 − 𝟑𝒙𝟐 − 𝟏𝟑𝒙 + 𝟏𝟓:
Calculations are found in the Appendix.
𝐴=(
3 − 4√3 128
3 + 4√3 128
,
,−
),𝐵 = (
) , 𝐺 = (1,0)
3
3
3√3
3√3
This gives us:
Equation
𝑦 = 𝑥 3 − 3𝑥 2 − 9𝑥 + 7
𝑦 = 𝑥 3 − 12𝑥 2 + 21𝑥 − 14
𝑦 = 𝑥 3 + 9𝑥 2 − 12
𝑦 = 𝑥 3 − 3𝑥 2 − 13𝑥 + 15
A
B
(−1, 12)
(1, −4)
(−6,96)
(
3 − 4√3 128
,
)
3
3√3
(
G
(3, −20)
(7, −122)
(0, −12)
(1, −4)
(4, −58)
(−3,42)
3 + 4√3
128
,−
)
3
3√3
(1,0)
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John McCarthy
Investigation of Cubic Functions
From these values, it appears that the value of G is the midpoint of A and B. Given the general values
A(𝑥1 , 𝑦1 ) and B(𝑥2 , 𝑦2 ), we can then use the midpoint formula, which would suggest that the
𝑥1 +𝑥2 𝑦1 +𝑦2
,
).
2
2
coordinates of the point of inflection are given by G(
Proof of the conjecture for a leading coefficient of 1:
In this section, we will attempt to prove the conjecture that the point of inflection for a cubic with a
leading coefficient of 1 is the midpoint of its two stationary points.
To find this, we must find the 𝑥 and 𝑦 values of the two stationary points and the 𝑥 and 𝑦 value of
the point of inflection, and then test these with the midpoint formula.
Given a general cubic polynomial with a leading coefficient of 1: 𝑦 = 𝑥 3 + 𝑝𝑥 2 + 𝑞𝑥 + 𝑟:
𝑑𝑦
= 3𝑥 2 + 2𝑝𝑥 + 𝑞
𝑑𝑥
𝑑2 𝑦
= 6𝑥 + 2𝑝
𝑑𝑥 2
𝑑𝑦
To find the 𝑥 values of the two stationary points, we must find when 𝑑𝑥 = 0.
3𝑥 2 + 2𝑝𝑥 + 𝑞 = 0
𝑥=
−2𝑝 ± √4𝑝2 − 4 × 3 × 𝑞 −𝑝 ± √𝑝2 − 3𝑞
=
6
3
Hence the stationary points occur at 𝑥 =
−𝑝±√𝑝2 −3𝑞
3
and 𝑥 =
−𝑝±√𝑝2 −3𝑞
.
3
𝑑2 𝑦
To find the 𝑥 value of the point of inflection we must find when 𝑑𝑥 2 = 0.
6𝑥 + 2𝑝 = 0
𝑥=−
𝑝
3
𝑝
3
Hence the point of inflection occurs at 𝑥 = − .
Substituting these 𝑥 values into the original equation, we can find the coordinates of the stationary
points and the point of inflection.
3
−𝑝+√𝑝2 −3𝑞
)
3
+𝑝(
2
−𝑝+√𝑝2 −3𝑞
)
3
+𝑞(
3
−𝑝−√𝑝2 −3𝑞
)
3
+𝑝(
2
−𝑝−√𝑝2 −3𝑞
)
3
+𝑞(
For 𝑥 =
−𝑝+√𝑝2 −3𝑞
:
3
𝑦=(
For 𝑥 =
−𝑝−√𝑝2 −3𝑞
:
3
𝑦=(
𝑝
3
𝑝 3
3
𝑝 2
3
𝑝
3
For 𝑥 = − : 𝑦 = (− ) + 𝑝 (− ) + 𝑞 (− ) + 𝑟 =
2 3
𝑝
27
−
−𝑝+√𝑝2 −3𝑞
)+
3
𝑟
−𝑝−√𝑝2 −3𝑞
)+
3
𝑟
𝑝𝑞
3
+𝑟
Using these coordinates for the two stationary points in the midpoint formula:
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John McCarthy
𝑚(𝑥, 𝑦) = (
Investigation of Cubic Functions
𝑥1 + 𝑥2 𝑦1 + 𝑦2
,
)
2
2
3
2
2
−𝑝 + √𝑝2 − 3𝑞 −𝑝 − √𝑝2 − 3𝑞 (−𝑝 + √𝑝 − 3𝑞) + ⋯ + 𝑞 (−𝑝 − √𝑝 − 3𝑞 ) + 𝑟
+
3
3
3
3
,
2
2
=
(
)
𝑝 2
𝑝𝑞
= (− , 𝑝3 −
+ 𝑟)
3 27
3
The result above is the coordinates of the point of inflection of the function, proving that the point
of inflection is the midpoint of the two stationary points.
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John McCarthy
Investigation of Cubic Functions
Investigating situations where this relationship does not apply:
In this section we will investigate situations where the values of the coefficients for the general
formula of a cubic where the leading coefficient is equal to 1 result in the above relationship not
applying.
As 𝑟 is a constant in this equation that is removed when
𝑑𝑦
𝑑𝑥
, and hence
𝑑2 𝑦
,
𝑑𝑥 2
are found, it has no
limits on the 𝑥 value of the point of stationary points or the point of inflection. However, 𝑟 does
affect the vertical translation of the cubic, as when 𝑥 values are substituted into the function, 𝑟 is
always added on to the corresponding 𝑦 coordinates.
As the roots of negative numbers are imaginary, and hence undefined in real numbers, for the
relationship to hold, 𝑝2 − 3𝑞 ≥ 0. Hence 𝑝2 ≥ 3𝑞.
If 𝑝2 < 3𝑞 then the function no longer has any stationary points, as the expression 𝑥 =
−𝑝±√𝑝2 −3𝑞
3
for the 𝑥 coordinates of the stationary points becomes undefined. However, the expression for the
𝑝
point of inflection, 𝑥 = − 3 , is unaffected, this suggests that in cubic polynomials where 𝑝2 < 3𝑞,
there is a single point of inflection but no stationary points. In the event that 𝑝2 = 3𝑞, the stationary
𝑝
3
point of the function would occur at 𝑥 = − , in which case the point would be a stationary point of
inflection.
An example of what the graphs would appear as for each of the above three cases:
𝒑𝟐 > 𝟑𝒒
𝒑𝟐 = 𝟑𝒒
𝒑𝟐 < 𝟑𝒒
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John McCarthy
Investigation of Cubic Functions
Investigating Cubic Functions where the leading coefficient is not equal to 𝟏:
The goal of this section is to investigate whether or not the relationship proposed above still holds
for cubic polynomials where the leading coefficient is not equal to 1. Again, letting A, B and G be the
two stationary points and the point of inflection respectively, we can find those points for the
following samples:
For 𝒚 = 𝟐𝒙𝟑 + 𝟏𝟐𝒙𝟐 + 𝟔𝒙 − 𝟏𝟒:
𝑑𝑦
= 6𝑥 2 + 24𝑥 + 6
𝑑𝑥
𝑑2 𝑦
= 12𝑥 + 24
𝑑𝑥 2
𝑑𝑦
To find the stationary points we must find when 𝑑𝑥 = 0.
6𝑥 2 + 24𝑥 + 6 = 0
𝑥=
−24 ± √242 − 4 × 6 × 6
= −2 ± √3
12
Hence 𝐴 = −2 − √3, 𝐵 = −2 + √3.
𝑑2 𝑦
To find the point of inflection we must find when 𝑑𝑥 2 = 0.
12𝑥 + 24 = 0
𝑥 = −2
Substituting the values of 𝑥 into the original equation, we find the coordinates:
𝐴 = (−2 − √3, 6 + 12√3), 𝐵 = (−2 + √3, 6 − 12√3), 𝐺 = (−2,6).
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John McCarthy
Investigation of Cubic Functions
For 𝒚 = 𝟑𝒙𝟑 + 𝟔𝒙𝟐 :
Calculations are found in the Appendix.
4 32
2 16
𝐴 = (− , ) , 𝐵 = (0,0) , 𝐺 = (− , )
3 9
3 9
For 𝒚 = 𝟐𝒙𝟑 + 𝟏𝟐𝒙𝟐 + 𝟏𝟖𝒙 + 𝟔:
Calculations are found in the Appendix.
𝐴 = (−3, 6), 𝐵 = (−1, −2), 𝐺 = (−2,2)
For 𝐲 = 𝟐𝐱^𝟑 + 𝟐𝟒𝐱:
Calculations are found in the Appendix.
𝐴 and 𝐵 do not exist, 𝐺 = (0, 0)
This gives us:
Equation
𝑦 = 2𝑥 3 + 12𝑥 2 + 6𝑥 − 14
𝑦 = 3𝑥 3 + 6𝑥 2
𝑦 = 2𝑥 3 + 12𝑥 2 + 18𝑥 + 6
y = 2x 3 + 24x
A
(−2 − √3, 6 + 12√3)
4 32
(− , )
3 9
(−3, 6)
𝑁/𝐴
B
(−2 + √3, 6 − 12√3)
(0,0)
(−1, −2)
𝑁/𝐴
G
(−2,6)
2 16
(− , )
3 9
(−2,2)
(0,0)
From these values, it appears again that the point of inflection is the midpoint of the two stationary
points, however this relationship does not hold true for the final equation, 𝑦 = 2𝑥 3 + 24𝑥, as there
are no stationary points for that function. We will investigate the reasons for this anomaly later.
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John McCarthy
Investigation of Cubic Functions
Proof of the conjecture for a leading coefficient not equal to 𝟏:
In this section we will attempt to prove the conjecture made above for cubic polynomials where the
leading coefficient is not equal to 1.
To prove this relationship, we must find when the 𝑥 or 𝑦 values of the two stationary points and the
𝑥 or 𝑦 values of the point of inflection and then compare these with the midpoint of the two
stationary points.
To attempt to prove this conjecture, we will use a general equation for a cubic polynomial:
𝑦 = 𝑎𝑥 3 + 𝑏𝑥 2 + 𝑐𝑥 + 𝑑
𝑑𝑦
= 3𝑎𝑥 2 + 2𝑏𝑥 + 𝑐
𝑑𝑥
𝑑2 𝑦
= 6𝑎𝑥 + 2𝑏
𝑑𝑥 2
𝑑𝑦
To find the 𝑥 coordinates of the two stationary points, we must find when 𝑑𝑥 = 0.
3𝑎𝑥 2 + 2𝑏𝑥 + 𝑐 = 0
−2𝑏 ± √4𝑏 2 − 4 × 3𝑎 × 𝑐 −𝑏 ± √𝑏 2 − 3𝑎𝑐
=
6𝑎
3𝑎
𝑥=
Hence the stationary points occur at 𝑥 =
−𝑏+√𝑏2 −3𝑎𝑐
3𝑎
and 𝑥 =
−𝑏−√𝑏2 −3𝑎𝑐
.
3𝑎
To find the 𝑥 coordinate of the point of inflection, we must find when
𝑑2 𝑦
𝑑𝑥 2
= 0.
6𝑎𝑥 + 2𝑏 = 0
𝑥=−
𝑏
3𝑎
𝑏
Hence the point of inflection occurs at 𝑥 = − 3𝑎.
Substituting the 𝑥 values above into the general equation 𝑦 = 𝑎𝑥 3 + 𝑏𝑥 2 + 𝑐𝑥 + 𝑑, we can attain
the 𝑦 coordinates for each point.
3
−𝑏+√𝑏2 −3𝑎𝑐
)
3𝑎
+𝑏(
2
−𝑏+√𝑏 2 −3𝑎𝑐
)
3𝑎
+𝑐(
3
−𝑏−√𝑏2 −3𝑎𝑐
)
3𝑎
+𝑏(
2
−𝑏−√𝑏 2 −3𝑎𝑐
)
3𝑎
+𝑐(
For 𝑥 =
−𝑏+√𝑏2 −3𝑎𝑐
:
3𝑎
𝑦 = 𝑎(
For 𝑥 =
−𝑏−√𝑏2 −3𝑎𝑐
:
3𝑎
𝑦 = 𝑎(
𝑏
𝑏
3
𝑏
2
𝑏
2𝑏3
−𝑏+√𝑏2 −3𝑎𝑐
)+
3𝑎
𝑑
−𝑏−√𝑏2 −3𝑎𝑐
)+
3𝑎
𝑑
𝑏𝑐
For 𝑥 = − 3𝑎: 𝑦 = 𝑎 (− 3𝑎) + 𝑏 (− 3𝑎) + 𝑐 (− 3𝑎) + 𝑑 = 27𝑎2 − 3𝑎 + 𝑑
Using the coordinates of the two stationary points in the midpoint formula:
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John McCarthy
Investigation of Cubic Functions
𝑚(𝑥, 𝑦)
3
2
2
−𝑏 + √𝑏 2 − 3𝑎𝑐 −𝑏 − √𝑏 2 − 3𝑎𝑐 𝑎 (−𝑏 + √𝑏 − 3𝑎𝑐) + ⋯ + 𝑐 (−𝑏 − √𝑏 − 3𝑎𝑐) + 𝑑
+
3𝑎
3𝑎
3𝑎
3𝑎
,
2
2
=
(
)
= (−
𝑏 2𝑏 3
𝑏𝑐
,
−
+ 𝑑)
2
3𝑎 27𝑎
3𝑎
The resulting coordinates are the same as the coordinates of the point of inflection, proving that for
all cubic polynomials of the form 𝑦 = 𝑎𝑥 3 + 𝑏𝑥 2 + 𝑐𝑥 + 𝑑, the point of inflection exists as the
midpoint of the two stationary points of the function.
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John McCarthy
Investigation of Cubic Functions
Investigating situations where this relationship does not apply:
In this section we will investigate situations where the coefficients of the 𝑥 terms of the general
function 𝑦 = 𝑎𝑥 3 + 𝑏𝑥 2 + 𝑐𝑥 + 𝑑 cause the above relationship to not hold. We have already
observed one function where this does not hold true, found above to be 𝑦 = 2𝑥 3 + 24𝑥. We will
investigate if the following restrictions explain why this function did not have any stationary points.
As 𝑑 is a constant value that does not exist as a coefficient, and hence is removed in the calculations
of the first and second derivatives of the cubic function, it has no effect on the 𝑥 values of the
stationary points or the point of inflection. However 𝑑 does effect the vertical translation of the
cubic polynomial, as it is added on to the corresponding 𝑦 value for any 𝑥 value of the function.
Given that the roots of negative numbers are imaginary, and hence undefined in real functions, 𝑏 2 −
3𝑎𝑐 ≥ 0, hence 𝑏 2 ≥ 3𝑎𝑐.
In the case that 𝑏 2 < 3𝑎𝑐, then the expressions for the stationary points of the function become
undefined, suggesting that the function has no stationary points, however the expression for the
𝑏
point of inflection for the function, 𝑥 = − 3𝑎 remains unaffected, resulting in the functions till having
a point of inflection even without stationary points. In the case that 𝑏 2 = 3𝑎𝑐, the stationary points
𝑏
of the function would both come to be the same value at − 3𝑎, in which case the function would
have a single stationary point of inflection.
There is also a limitation on the value of 𝑎, in that 𝑎 ≠ 0, as this would cause the point of inflection
to be undefined. However, as this would effectively make the function a quadratic, for the purposes
of this investigation the effects of this can be ignored.
Looking back at the equation found earlier that did not abide by the relationship found, 𝑦 = 2𝑥 3 +
24𝑥, we see that in this equation, 𝑏 = 0, and hence 𝑏 2 < 3𝑎𝑐. In the above investigation, we found
that when this is the case, the function has no stationary points, which is consistent with the results
found for the stationary points of this function, proving that it does not invalidate the relationships
found above.
An example of what a graph of a cubic would like for the above three cases:
𝒃𝟐 > 𝟑𝒂𝒄
𝒃𝟐 = 𝟑𝒂𝒄
𝒃𝟐 < 𝟑𝒂𝒄
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John McCarthy
Investigation of Cubic Functions
Conclusion:
During this investigation, we have found the relationships between the two stationary points of a
cubic polynomial and its point of inflection, in the case of the cubic polynomial having, or not having,
a leading coefficient equal to 1. The relationship found is that the coordinates of the point of
inflection are the midpoint of the two stationary points of the cubic function. We have also found
the cases where this relationship does not apply, including when there is only a single stationary
point that exists as a stationary point of inflection, and when there are no stationary points.
Page 11