The Cubic Curve 2

Years

Year 10

Graphs

The Cubic Curve

The Cubic Curve 2

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Question 1 of 4

1. Question
Write the equation of the cubic graph that is shown on the number plane below.
- 1. $y = 2 x^{3} - 4$ $y=2x^3-4$
- 2. $y = 4 x^{3} - 2$ $y=4x^3-2$
- 3. $y = \frac{3}{2} x^{3}$ $y=\frac{3}{2}x^3$
- 4. $y = x^{3} - 4$ $y=x^3-4$
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This cubic curve does not have any turning points and it passes through the origin. Therefore it is in the form $y = a x^{3}$ $y=ax^3$ .

The value of $d$ $d$ will be given by the $y$ $y$ -intercept.

The graph crosses the $y$ $y$ -axis at $(0,$ $(0,$ $- 4$ $-4$ $)$ $)$ , therefore $d = - 4$ $d=-4$ .

The equation of the graph has the form $y = a x^{3}$ $y=ax^3$ $- 4$ $-4$ .

Use the point $(2, 12)$ $(2,12)$ shown on the graph to find the value of $a$ $a$ .

The $x$ $x$ -coordinate of this point is $2$ $2$ , and the $y$ $y$ -coordinate of this point is $12$ $12$ .

So we should substitute $x = 2$ $x=2$ and $y = 12$ $y=12$ into the equation $y$ $y$ $= a$ $=a$ $x^{3}$ $x^3$ $- 4$ $-4$ .

$12$ $12$ $= a$ $=a$ $(2^{3})$ $(2^3)$ $- 4$ $-4$ .

Solve this equation to find the value of $a$ $a$ .

$12 = 8 a - 4$ $12=8a-4$

$16 = 8 a$ $16=8a$

$\frac{16}{8} = a$ $\frac{16}{8}=a$

$a = 2$ $a=2$

Substitute the value for $a$ $a$ and write the equation of the graph.

$y = 2 x^{3} - 4$ $y=2x^3-4$
Question 2 of 4

2. Question
Write the equation of the cubic graph that is shown on the number plane below.
- 1. $y = - \frac{1}{3} x^{3} + 2$ $y=-\frac{1}{3}x^3+2$
- 2. $y = \frac{1}{3} x^{3} - 2$ $y=\frac{1}{3}x^3-2$
- 3. $y = 3 x^{3} + 2$ $y=3x^3+2$
- 4. $y = x^{3} + 2$ $y=x^3+2$
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This cubic curve does not have any turning points and it passes through the origin. Therefore it is in the form $y = a x^{3}$ $y=ax^3$ .

The value of $d$ $d$ will be given by the $y$ $y$ -intercept.

The graph crosses the $y$ $y$ -axis at $(0,$ $(0,$ $2$ $2$ $)$ $)$ , therefore $d = 2$ $d=2$ .

The equation of the graph has the form $y = a x^{3}$ $y=ax^3$ $+ 2$ $+2$ .

Use the point $(3, - 7)$ $(3,-7)$ shown on the graph to find the value of $a$ $a$ .

The $x$ $x$ -coordinate of this point is $3$ $3$ , and the $y$ $y$ -coordinate of this point is $- 7$ $-7$ .

So we should substitute $x = 3$ $x=3$ and $y = - 7$ $y=-7$ into the equation $y$ $y$ $= a$ $=a$ $x^{3}$ $x^3$ $+ 2$ $+2$ .

$- 7$ $-7$ $= a$ $=a$ $(3^{3})$ $(3^3)$ $+ 2$ $+2$ .

Solve this equation to find the value of $a$ $a$ .

$- 7 = 27 a + 2$ $-7=27a+2$

$- 9 = 27 a$ $-9=27a$

$- \frac{27}{9} = a$ $-\frac{27}{9}=a$

$a = - \frac{1}{3}$ $a=-\frac{1}{3}$

Substitute the value for $a$ $a$ and write the equation of the graph.

$y = - \frac{1}{3} x^{3} + 2$ $y=-\frac{1}{3}x^3+2$
Question 3 of 4

3. Question
The following equation is graphed below:
$y = a (x - 1) (x + 1) (x - 3)$ $y=a(x-1)(x+1)(x-3)$

What is the value of $a$ $a$ ?
- 1. $- \frac{9}{2}$ $-\frac{9}{2}$
- 2. $\frac{3}{2}$ $\frac{3}{2}$
- 3. $- 3$ $-3$
- 4. $3$ $3$
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The standard form for graphing a cubic is $y = a (x - r) (x - s) (x - t)$ $y=a(x-r)(x-s)(x-t)$ .

Since the point $($ $($ $2$ $2$ $,$ $,$ $- 9$ $-9$ $)$ $)$ lies on the graph we can substitute $x = 2$ $x=2$ and $y = - 9$ $y=-9$ into the equation and solve to find $a$ $a$ .

$- 9$ $-9$ $= a ($ $=a($ $2$ $2$ $- 1) ($ $-1)($ $2$ $2$ $+ 1) ($ $+1)($ $2$ $2$ $- 3)$ $-3)$

$- 9 = a (1) (3) (- 1)$ $-9=a(1)(3)(-1)$

$- 9 = - 3 a$ $-9=-3a$

$\frac{- 9}{- 3} = a$ $\frac{-9}{-3}=a$

$a = 3$ $a=3$
Question 4 of 4

4. Question
Write the equation of the cubic graph shown below.
- 1. $y = - 3 (x - 1) (x + 2) (x - 2)$ $y=-3(x-1)(x+2)(x-2)$
- 2. $y = 3 (x - 1) {(x - 2)}^{2}$ $y=3(x-1)(x-2)^2$
- 3. $y = 3 {(x - 1)}^{2} (x + 2)$ $y=3(x-1)^2(x+2)$
- 4. $y = - 3 (x - 1) {(x + 2)}^{2}$ $y=-3(x-1)(x+2)^2$
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The standard form for graphing a cubic is $y = a (x - r) (x - s) (x - t)$ $y=a(x-r)(x-s)(x-t)$ , where $r$ $r$ , $s$ $s$ , and $t$ $t$ are the $x$ $x$ -intercepts of the graph.

Find the values of the $x$ $x$ -intercepts of the graph.

One $x$ $x$ -intercept is at $1$ $1$ , so $r = 1$ $r=1$ .

The graph touches the $x$ $x$ -axis at $- 2$ $-2$ , meaning it will be a repeated root, so $s = - 2$ $s=-2$ and $t = - 2$ $t=-2$

Substitute the values for $r$ $r$ , $s$ $s$ and $t$ $t$ into the standard equation

$y = a (x -$ $y=a(x-$ $1$ $1$ $) (x - ($ $)(x-($ $- 2$ $-2$ $)) (x - ($ $))(x-($ $- 2$ $-2$ $))$ $))$

$y = a (x - 1) {(x + 2)}^{2}$ $y=a(x-1)(x+2)^2$

Since the graph passes through $($ $($ $0$ $0$ $,$ $,$ $12$ $12$ $)$ $)$ , substitute $x = 0$ $x=0$ and $y = 12$ $y=12$ into the equation and solve to find $a$ $a$ .

$12$ $12$ $= a ($ $=a($ $0$ $0$ $- 1) ($ $-1)($ $0$ $0$ $+ 2)^{2}$ $+2)^2$

$12 = a (- 1) {(2)}^{2}$ $12=a(-1)(2)^2$

$12 = - 4 a$ $12=-4a$

$\frac{12}{- 4} = a$ $\frac{12}{-4}=a$

$a = - 3$ $a=-3$

Substitute the value for $a$ $a$ and write the complete equation.

$y = - 3 (x - 1) {(x + 2)}^{2}$ $y=-3(x-1)(x+2)^2$