Graph Cubic Curves

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Question 1 of 2

1. Question
Sketch the graph of the cubic curve for the equation below

$y = 2 (x + 1) (x - 4) (x - 6)$ $y=2(x+1)(x-4)(x-6)$
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The standard form for graphing a cubic is $y = a (x - r) (x - s) (x - t)$ $y=a(x-r)(x-s)(x-t)$ .

The sign of $a$ $a$ in this equation tells us the shape of the graph

In this equation $a = 2$ $a=2$ which is positive.

Therefore this is a positive cubic curve and has the approximate shape shown below:

Find the $x$ $x$ -intercepts of the graph by setting $y = 0$ $y=0$ .

$2$ $2$ $(x + 1)$ $(x+1)$ $(x - 4)$ $(x-4)$ $(x - 6)$ $(x-6)$ $= 0$ $=0$

So $x + 1 = 0$ $x+1=0$ , $x - 4 = 0$ $x-4=0$ or $x - 6 = 0$ $x-6=0$

This means $x = - 1$ $x=-1$ , $x = 4$ $x=4$ or $x = 6$ $x=6$ .

This gives us $x$ $x$ -intercepts $- 1$ $-1$ , $4$ $4$ and $6$ $6$ which we mark on the $x$ $x$ -axis

Find the $y$ $y$ -intercepts of the graph by setting $x = 0$ $x=0$ .

$y = 2 (0 + 1) (0 - 4) (0 - 6)$ $y=2(0+1)(0-4)(0-6)$

$y = 2 \cdot (1) \cdot (- 4) \cdot (- 6)$ $y=2\cdot (1) \cdot (-4)\cdot (-6)$

$y = 48$ $y=48$

This gives us a $y$ $y$ -intercept of $48$ $48$ which we mark on the $y$ $y$ -axis

Draw a positive cubic curve which passes through the intercepts found.
Question 2 of 2

2. Question
Sketch the graph of the cubic curve for the equation below

$y = - 2 (x - 1) (x + 2) (x + 4)$ $y=-2(x-1)(x+2)(x+4)$
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The standard form for graphing a cubic is $y = a (x - r) (x - s) (x - t)$ $y=a(x-r)(x-s)(x-t)$ .

The sign of $a$ $a$ in this equation tells us the shape of the graph

In this equation $a = - 2$ $a=-2$ which is negative.

Therefore this is a negative cubic curve and has the approximate shape shown below:

Find the $x$ $x$ -intercepts of the graph by setting $y = 0$ $y=0$ .

$- 2$ $-2$ $(x - 1)$ $(x-1)$ $(x + 2)$ $(x+2)$ $(x + 4)$ $(x+4)$ $= 0$ $=0$

So $x - 1 = 0$ $x-1=0$ , $x + 2 = 0$ $x+2=0$ or $x + 4 = 0$ $x+4=0$

This means $x = 1$ $x=1$ , $x = - 2$ $x=-2$ or $x = - 4$ $x=-4$ .

This gives us $x$ $x$ -intercepts $1$ $1$ , $- 2$ $-2$ and $- 4$ $-4$ which we mark on the $x$ $x$ -axis

Find the $y$ $y$ -intercepts of the graph by setting $x = 0$ $x=0$ .

$y = - 2 (0 - 1) (0 + 2) (0 + 4)$ $y=-2(0-1)(0+2)(0+4)$

$y = - 2 \cdot (- 1) \cdot (2) \cdot (4)$ $y=-2\cdot (-1) \cdot (2)\cdot (4)$

$y = 16$ $y=16$

This gives us a $y$ $y$ -intercept of $16$ $16$ which we mark on the $y$ $y$ -axis

Draw a negative cubic curve which passes through the intercepts found.

Graph Cubic Curves

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