Graphs

What can we find out from the function

itself?

€

f (x) = x 3 + 3x 2 −13x −15

€

f (x) = x 3 + 3x 2 −13x −15 = 0

Take the function

To find the roots

€

f (x) = (x + 5)(x +1)(x − 3) = 0

€

x = −5, −1, 3

Function

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-6 -4 -2 0 2 4

-5 -1 3€

f (x) = x 3 + 3x 2 −13x −15

Stationary Points

€

f (x) = x 3 + 3x 2 −13x −15

Find where the first derivative is zero

€

′ f (x) = 3x 2 + 6x −13 = 0

€

x =1.31, − 3.31

Substitute x-values to find y-values

€

y = −24.6, 24.6

(1.31, -24.6), (-3.31, 24.6)

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€

f (x)

€

′ f (x)

(1.31, -24.6)

(-3.31, 24.6)

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(1.31, -24.6)

(-3.31, 24.6)

Gradient function is positive i.e.

€

′ f (x) > 0

Function is increasing

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-6 -4 -2 0 2 4

(1.31, -24.6)

(-3.31, 24.6)

Gradient function is positive i.e.

€

′ f (x) > 0

Function is increasing

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-6 -4 -2 0 2 4

(1.31, -24.6)

(-3.31, 24.6)

Gradient function is negative i.e.

€

′ f (x) < 0

Function is decreasing

Nature of turning points

€

f (x) = x 3 + 3x 2 −13x −15

€

′ f (x) = 3x 2 + 6x −13 = 0

FunctionFirst derivativeSecond derivative

€

′ ′ f (x) = 6x + 6

Substitute the x-values of the stationary points €

′ ′ f (1.31) =13.9

€

′ ′ f (−3.31) = −13.86

Positive indicates minimum

Negative indicates maximum

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€

f (x)

€

′ f (x)

€

′ ′ f (x)

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is a maximum

€

f (x)

is negative

€

′ ′ f (x) is a minimum

€

f (x)

is positive

€

′ ′ f (x)

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is concave down

€

f (x)

is negative

€

′ ′ f (x)

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is concave up

€

f (x)

is positive

€

′ ′ f (x)

• Concave Up - 2nd derivative positive

• Concave Down - 2nd derivative negative

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has a point of inflection

€

f (x)

is zero

€

′ ′ f (x)There is a change

in curvature

Example 1

Find the stationary points of the following function and determine their nature.

€

y = x 5 + 5x 4 + 5x 3 +1

To find the roots

Roots are:

(-3.63, 0) (-1, 0)

€

y = x 5 + 5x 4 + 5x 3 +1 = 0

Using solver on graphics calculator

€

x=−3.63, −1

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yx = -3.63

Example 1

To find the stationary points.

€

y = x 5 + 5x 4 + 5x 3 +1

Differentiate

€

dy

dx= 5x 4 + 20x 3 +15x 2 = 0

Factorise

€

5x 2 x 2 + 4x + 3( ) = 5x 2 x + 3( ) x +1( ) = 0

Stationary Points are:

(0, 1), (-1, 0), (-3, 28)

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y

-3, 28

-1, 0

0, 1

€

′ f (x)-30

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x

y

The first derivative tells us where the function is increasing/decreasing and where it is stationary.

€

f (x)

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x

y

The first derivative tells us where the function is increasing/decreasing and where it is stationary.

€

x = −3

€

x = −1

′ f (x) = 0

€

x = 0

′ f (x) = 0

Function is stationary

€

′ f (x) = 0

Function is stationary

Function is stationary

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x

y

The first derivative tells us where the function is increasing/decreasing and where it is stationary.

Gradient is positive

€

′ f (x) > 0

€

x < −3

€

−1< x < 0

′ f (x) > 0

€

x > 0

′ f (x) > 0

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x

y

The first derivative tells us where the function is increasing/decreasing …

€

x < −3

€

−1< x < 0

′ f (x) > 0

€

x > 0

′ f (x) > 0

Function is increasing

€

′ f (x) > 0

Function is increasing

Function is increasing

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x

y

The first derivative tells us where the function is increasing/decreasing …

€

′ f (x) < 0

Function is decreasing

To determine the nature of the turning

points:Differentiate again:

€

d2y

dx 2= 20x 3 + 60x 2 + 30x

€

′ ′ f (−1) = +ve ⇒ min

′ ′ f (−3) = −ve ⇒ max

′ ′ f (0) = 0⇒ investigate gradients each side of 0

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y

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′ ′ f (x)

€

f (x)

€

′ f (x)

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y

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′ ′ f (x) < 0

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f (x)x = -3

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y

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′ ′ f (x) > 0

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f (x)x = -1

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y

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′ ′ f (x) = 0

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f (x)x = 0

Let’s take a closer look!

-5

5

-0.5 0.5x

y

x = 0

€

′ ′ f (x) = 0

€

f (x)

This means we need to look at the gradient

function.

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5

-0.5 0.5x

y

x = 0

€

′ ′ f (x) = 0

€

f (x)

Before ‘0’, the gradient is negative.

-5

5

-0.5 0.5x

y

x = 0

€

′ ′ f (x) = 0

€

f (x)

After ‘0’, the gradient is positive.

To determine the nature of the turning

points:Differentiate again:

€

d2y

dx 2= 20x 3 + 60x 2 + 30x

€

′ ′ f (−1) = +ve ⇒ min

′ ′ f (−3) = −ve ⇒ max

′ ′ f (0) = 0⇒ investigate gradients each side of 0

Gradient is negative just before “0” and positive just after “0” minimum

€

⇒

Practice: Concavity

Find where the following function is concave down.

€

y = x 3 +12x 2 + 48x + 6

Differentiate twice:

€

dy

dx= 3x 2 + 24 x + 48

d2y

dx 2= 6x + 24 < 0

€

⇒ x < −4

Practice: Find where the function is increasing

€

y = x 3 − 3x 2 − 9x + 5

€

dy

dx= 3x 2 − 6x − 9 > 0⇒ 3(x − 3)(x +1) > 0

Draw the graph

€

x < −1, x > 3

Topic 11 Reference Page Exercise

Delta167,172

13.4, 13.5,13.6

Sidebotham160,165

8.1, 8.2

NuLake47,

53, 581.13, 1.14,

1.15Theta (red) 84, 87 8.3, 8.4

• locate maxima and minima of polynomialfunctions

• sketch graphs of polynomials of degree≥3 and identifying features (turningpoints, points of inflection and concavity).

SeniorMathematics

201 1A, 1B