The cubic function belongs to the family of polynomial functions. The image below summarises the different shapes of cubics that you might encounter. The black circle is indicating the point of inflection on each photo. It is interesting to see that the cubics in photo A have turning points, called a local maximum or minimum, while the other cubics do not. While we don't need to use the terms written on the photos right now, they will reappear when we study Calculus later on.
Just as with linear and quadratic functions, cubics can be represented in different forms. For this course there are three main forms that we need to be familiar with:
General Form: $y=ax^3+bx^2+cx+d$y=ax3+bx2+cx+d
Translated form: $y=k\left(x-b\right)^3+c$y=k(x−b)3+c
Factored form ($3$3 linear factors): $y=k\left(x-a\right)\left(x-b\right)\left(x-c\right)$y=k(x−a)(x−b)(x−c)
Depending on the cubic involved, the function could be a one-to-one relation or a many-to-one relation. (Recall that a one-to-one relation is a relation where no two values of $x$x are mapped to the same value of $y$y, and a many-to-one relation is a relation where two or more values of $x$x are mapped to the same value of $y$y).
Given the general form of a cubic function $y=ax^3+bx^2+cx+d$y=ax3+bx2+cx+d, various combinations of the four coefficients $a$a, $b$b, $c$c and $d$d will create variations in the shape and position of the curve. The applet below gives us a chance to experiment with this.
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Although cubics can vary in their shape, there are some common features:
This second applet allows you to investigate the form translated form depicted in photo B above. The translated form is expressed as $y=k\left(x-b\right)^3+c$y=k(x−b)3+c . This form represents the curve of the function $y=kx^3$y=kx3 translated (shifted in position without distorting the shape) horizontally $b$b units and vertically $c$c units.
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$y=k\left(x-a\right)\left(x-b\right)\left(x-c\right)$y=k(x−a)(x−b)(x−c)
Just like factored form for quadratics, this form allows us to read off the $x$x-intercepts directly. Try changing the values in the applet below, try cases with all 3 values different and other cases where some or all of the values match.
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For the three forms above the $y$y-intercept can be found by making $x=0$x=0. Or if you have the general form the constant term $d$d is the point where the function cuts the $y$y-axis.
When finding the $x$x-intercepts the general form can be difficult to use with the skills we have developed so far, while the translated form can be solved by making $y=0$y=0. The $x$x-intercepts for the factored form can be determined directly from the equation as described above or solved in a similar way to a quadratic equation using the null factor law.
Find the $x$x-intercepts of the function $y=2(x-3)^3-16$y=2(x−3)3−16.
Think: The above equation is in the translated form so we can find the $x$x-intercept by solving for $y=0$y=0.
Do: Set $y$y to $0$0 and solve for $x$x
$2(x-3)^3-16$2(x−3)3−16 | $=$= | $0$0 |
$2(x-3)^3$2(x−3)3 | $=$= | $16$16 |
$(x-3)^3$(x−3)3 | $=$= | $8$8 |
$x-3$x−3 | $=$= | $2$2 |
$x$x | $=$= | $5$5 |
Find the $x$x-intercepts of the function $y=2(x-1)(x-3)(x+2)$y=2(x−1)(x−3)(x+2).
Think: The $x$x-intercepts occur at $y=0$y=0. The equation is in factored form though so we can use the null factor law to read off the $x$x-intercepts.
Do: Set the value of $y$y to $0$0 and read off the $y$y-intercepts.
$2(x-1)(x-3)(x+2)$2(x−1)(x−3)(x+2) | $=$= | $0$0 |
$(x-1)(x-3)(x+2)$(x−1)(x−3)(x+2) | $=$= | $0$0 |
$x$x | $=$= | $-2,1,3$−2,1,3 using the null factor law |
Consider the graph of the function.
For what values of $x$x is the cubic concave up?
For what values of $x$x is the cubic concave down?
State the coordinates of the point of inflection in the form $\left(a,b\right)$(a,b).
Consider the given graph of a cubic function.
Determine whether the cubic is positive or negative.
negative
positive
State the coordinates of the $y$y-intercept.
Which of the following could be the equation of the function?
$y=-2x^3+3$y=−2x3+3
$y=2x^3+3$y=2x3+3
$y=2x^3-3$y=2x3−3
$y=-2x^3-3$y=−2x3−3
Consider the curve $y=-3\left(x-1\right)^3+3$y=−3(x−1)3+3.
Find the value of the $x$x-intercept.
Find the value of the $y$y-intercept.
State the coordinates of the point of inflection.
Now use the point of inflection to draw a graph of the curve.
Consider the curve $y=\left(x+3\right)\left(x+2\right)\left(x-2\right)$y=(x+3)(x+2)(x−2).
Find the $x$x-value of the $x$x-intercept(s).
Find the $y$y-value of the $y$y-intercept(s).
Sketch a graph of the curve.
Which of the following functions has a curve that passes through the origin?
$y=\left(x-2\right)^2\left(x+3\right)$y=(x−2)2(x+3)
$y=\left(x+1\right)^3$y=(x+1)3
$y=\left(x-4\right)\left(x+7\right)\left(x-5\right)$y=(x−4)(x+7)(x−5)
$y=x\left(x-6\right)\left(x+8\right)$y=x(x−6)(x+8)