02.14cubicgraphs

# Cubic Graphs

The general form of the cubic function is $y = ax^3 + bx^2 + cx + d$

the sign of a (the coefficient of x3) controls the direction of the curve

• a > 0, the graph trends up from left to right (ie it starts in 3rd Quadrant, ends in 1st Quadrant)
• a < 0, the graph trends down from left to right (ie it starts in 2nd Quadrant, ends in 4th Quadrant)

The constant term (d) gives the y-intercept

Some cubics have a single stationary point of inflection, the rest have two turning points (a local minimum and a local maximum)

A cubic may have one or two or three x-intercepts

• The turning points are NOT midway between the x-intercepts
• The x-intercepts are given by the solutions of $ax^3 + bx^2+ cx + d = 0$
• This can be solved by factorising, then using the Null Factor Law
• Factors can be found using the Factor Theorem {If P(a) = 0, then x – a is a factor}
• A CAS calculator can factorise {select factor from the ACTION menu, TRANSFORMATION submenu}
• A CAS calculator can solve the equation P(x) = 0 to find the x-intercepts

The x-value of the stationary points can be found by

• setting the derivative to zero and solving for x
• graphing on a CAS calcuator {Max and Min are in the ANALYSIS menu, G-SOLVE submenu}

# Basic Form (or Power Form)

Some cubics are transformations of the basic cubic, $y = x^3$.

The basic cubic has a stationary point of inflection at the origin (no turning points).

The Power Form of the cubic can be written as:

… … $y = a(x - h)^3 + k$

The usual transformations apply:

The sign of a controls the direction of the curve

• a > 0, the graph trends up from left to right
• a < 0, the graph trends down from left to right

The magnitude of a controls the dilation of the curve

• a > 1, (or a < –1), the graph is thinner
• a < 1, (or a > –1), the graph is wider

The stationary point of inflection is at (x = h, y = k)

To find the y-intercept, substitute x = 0
To find the x-intercept (there will be exactly one), substitute y = 0

#### Example 1

… Sketch $y = 2(x - 2)^3 - 1$

Solution:

… … Stationary point at (2, –1)
… … Dilation by a factor of 2 (thinner)
… … a > 0 so trend is up from left to right

Y-intercept

… … $y = 2(-2)^3 - 1$

… … $y = -17$

X-intercept

… … $2 \big( x - 2 \big)^3 - 1 = 0$

… … $2 \big( x - 2 \big)^3 = 1$

… … $(x - 2)^3 = \dfrac{1}{2}$

… … $x-2=\sqrt[3]{\dfrac{1}{2}}$

… … $x=2+\sqrt[3]{\dfrac{1}{2}}$

… … $x \approx 2.8$

## Factor Form

Some cubic equations can be factorised into the form:

… … $y = a(x - b)(x - c)(x - d)$

the sign of a controls the direction of the curve

• +a trends up from left to right
• –a trends down from left to right

the magnitude of a gives the dilation factor

b, c and d are the x-intercepts

• a factor repeated twice indicates a turning point on the axis at that x-value

to find the y-intercept, substitute x = 0

the turning points are NOT midway between the x-intercepts

#### Example 2

… … Sketch $y = 0.5(x + 3)(x - 1)(x - 4)$

Solution:

X-intercepts

… … From the equation, $x = -3, \; 1, \; 4$

Y-intercept

… … substitute $x = 0$

… … $y = 0.5(3)(-1)(-4)$

… … $y = 6$

Dilation Factor

… … $a = 0.5$ (wider)

… … a > 0 so it is a positive cubic, trends up from left to right

Stationary Points

… … Sketch graph on your calculator, then use it to get values for the turning points

… … … {Max and Min are in the ANALYSIS menu, G-SOLVE submenu}

… … Local maximum at (–1.36, 10.37)

… … Local minimum at (2.69, –6.30)

#### Example 3

… … Sketch $y = -(x + 2)(x - 1)^2$

Solution:

X-intercepts

… … From the equation $x = -2, \; 1$

Y-intercept

… … Substitute $x = 0$

… … $y = -(2)(-1)^2$

… … $y = -2$

Dilation Factor

… … $a = -1$ (dilation factor = 1, standard width)

… … a < 0 so negative cubic, trends down

Stationary Points

… … Repeated factor means one turning point is on the axis at x = 1

… … Use calculator for the other turning point

… … Local maximum at (1, 0)

… … Local minimum at (–1, –4)

Note:

• Stationary points can be located algebraically by
• Finding the derivative of the function
• Setting the derivative to equal zero

## Cubics with restricted domains

If the domain of a cubic is R, then the range is also R

If the domain is restricted, then care must be taken when stating the range

• the endpoints for the domain may not give the maximum or minimum values for the range.
• the y-values of the turning points must also be considered.

#### Example 4

… … Sketch $y = \frac{1}{3}x^3 + x^2 -3x-3 \text{ in the domain } x \in \big( -5\frac{1}{4}, \,2 \big]$ and state the range

Solution:

• With the aid of a calculator, draw the entire graph (in pencil),
• then mark in the endpoints and draw the desired section more clearly.
• Erase the unwanted portions of the graph.

X-intercepts

… … x = –4.54, –0.83, 2.38

Y-intercepts

… … y = –3

Turning Points

… … Local maximum (–3, 6)

… … Local minimum (1, –4.67)

Endpoints

… … (–5.25, –8) … {open circle}

… … (2, –2.33) … {closed circle}

Desired section is between the two endpoints (in dark blue)

Domain: … $x \in (-5.25, \; 2]$

Range: … $y \in (-8, \; 6]$ … by observation of the graph

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