Transformations of Cubic and Quartic Functions

In Grade 11 you transformed parabolas, square roots, and reciprocals. Exactly the same rules work for the power functions $y = x^3$ and $y = x^4$. Once you know the shapes of these two parents, you can sketch something like $y = -\tfrac{1}{2}(x - 1)^3 + 3$ in under a minute.

Key ideas

The two parents

	$y = x^3$	$y = x^4$
key points	$(-2, -8)$, $(-1, -1)$, $(0, 0)$, $(1, 1)$, $(2, 8)$	$(-2, 16)$, $(-1, 1)$, $(0, 0)$, $(1, 1)$, $(2, 16)$
shape	flattens through the origin, Q3 to Q1	U-shaped but flatter at the bottom than $y = x^2$, Q2 to Q1
special point	point of symmetry $(0, 0)$	vertex $(0, 0)$
range	$\{y \in \mathbb{R}\}$	$\{y \in \mathbb{R} \mid y \ge 0\}$

The parameters

$$y = a\,f\big(k(x - d)\big) + c$$

As in combining transformations:

Parameter	Effect
$a$	vertical stretch or compression by a factor of $\lvert a \rvert$; reflection in the $x$-axis if $a \lt 0$
$k$	horizontal stretch or compression by a factor of $\dfrac{1}{\lvert k \rvert}$; reflection in the $y$-axis if $k \lt 0$
$d$	horizontal translation: right if $d \gt 0$, left if $d \lt 0$
$c$	vertical translation: up if $c \gt 0$, down if $c \lt 0$

The mapping rule is

$$(x, y) \to \left(\frac{x}{k} + d,\ ay + c\right)$$

Where the special point goes

The point $(0, 0)$ maps to $(d, c)$. So:

• for a cubic $y = a(x - d)^3 + c$, the point of symmetry is $(d, c)$, and the range is always $\{y \in \mathbb{R}\}$;
• for a quartic $y = a(x - d)^4 + c$, the vertex is $(d, c)$. The range is $\{y \in \mathbb{R} \mid y \ge c\}$ if $a \gt 0$, or $\{y \in \mathbb{R} \mid y \le c\}$ if $a \lt 0$.

The domain is $\{x \in \mathbb{R}\}$ in every case.

Worked examples

Example 1: Describing a quartic

Describe the transformations of $y = x^4$ in $y = 3(x + 2)^4 - 5$. Give the vertex, the direction of opening, and the range.

Solution. $a = 3$, $k = 1$, $d = -2$, $c = -5$.

• Vertical stretch by a factor of $3$.
• Translation $2$ units left and $5$ units down.

The vertex is $(-2, -5)$. Since $a \gt 0$, the graph opens up, so $-5$ is the minimum value.

Range: $\{y \in \mathbb{R} \mid y \ge -5\}$.

Example 2: Sketching a cubic

Sketch $y = -\tfrac{1}{2}(x - 1)^3 + 3$. State its point of symmetry and $y$-intercept.

Solution. $a = -\tfrac{1}{2}$, $k = 1$, $d = 1$, $c = 3$: a vertical compression by a factor of $\tfrac{1}{2}$, a reflection in the $x$-axis, and a translation $1$ right and $3$ up. The mapping rule is $(x, y) \to \left(x + 1,\ -\tfrac{1}{2}y + 3\right)$.

$y = x^3$	$(-2, -8)$	$(-1, -1)$	$(0, 0)$	$(1, 1)$	$(2, 8)$
image	$(-1, 7)$	$(0, 3.5)$	$(1, 3)$	$(2, 2.5)$	$(3, -1)$

For example, $(-2, -8) \to \left(-2 + 1,\ -\tfrac{1}{2}(-8) + 3\right) = (-1, 7)$.

The reflection makes the cubic fall from left to right; it flattens out at $(1, 3)$.

The point of symmetry is $(1, 3)$. The $y$-intercept is $3.5$ (check: $-\tfrac{1}{2}(0 - 1)^3 + 3 = \tfrac{1}{2} + 3 = 3.5$ ✓).

Example 3: Factoring out k

Describe the transformations in $y = (2x + 6)^4 - 1$, map the key points of $y = x^4$, and find the $x$-intercepts.

Solution. Factor first: $y = \big(2(x + 3)\big)^4 - 1$. So $a = 1$, $k = 2$, $d = -3$, $c = -1$.

• Horizontal compression by a factor of $\tfrac{1}{2}$.
• Translation $3$ units left and $1$ unit down.

The mapping rule is $(x, y) \to \left(\tfrac{x}{2} - 3,\ y - 1\right)$:

$y = x^4$	$(-2, 16)$	$(-1, 1)$	$(0, 0)$	$(1, 1)$	$(2, 16)$
image	$(-4, 15)$	$(-3.5, 0)$	$(-3, -1)$	$(-2.5, 0)$	$(-2, 15)$

The horizontal compression makes the graph narrow; its vertex is $(-3, -1)$.

Two of the mapped points land on the $x$-axis, so the $x$-intercepts are $-3.5$ and $-2.5$. Check algebraically:

$$(2x + 6)^4 = 1 \quad\Rightarrow\quad 2x + 6 = \pm 1 \quad\Rightarrow\quad x = -2.5 \ \text{ or } \ x = -3.5$$

(Since $\big(2(x + 3)\big)^4 = 16(x + 3)^4$, you could also see this graph as a vertical stretch by $16$. Both descriptions give the same graph.)

Example 4: Finding the equation

A cubic of the form $y = a(x - d)^3 + c$ has point of symmetry $(1, -2)$ and passes through $(2, 1)$. Find its equation.

Solution. The point of symmetry gives $d = 1$ and $c = -2$:

$$y = a(x - 1)^3 - 2$$

Substitute $(2, 1)$:

$$1 = a(2 - 1)^3 - 2 \quad\Rightarrow\quad 1 = a - 2 \quad\Rightarrow\quad a = 3$$

The equation is $y = 3(x - 1)^3 - 2$.

Common mistakes

Getting the direction of the horizontal shift wrong. $y = (x + 2)^4$ is $2$ units left, since $d = -2$.

Not factoring out $k$. In $y = (2x + 6)^4$, the shift is $3$ left, not $6$. Rewrite as $\big(2(x + 3)\big)^4$ first.

Translating before stretching. For the $y$-coordinate, multiply by $a$ first, then add $c$. Under $y = -\tfrac{1}{2}(x - 1)^3 + 3$, the point $(2, 8)$ goes to $y = -\tfrac{1}{2}(8) + 3 = -1$, not $-\tfrac{1}{2}(8 + 3)$.

Giving a cubic a maximum or minimum. A transformed $y = x^3$ still goes from one end to the other forever, so its range is all real numbers. Only the quartic has a vertex with a max or min value.

Forgetting what a reflection does to a quartic’s range. If $a \lt 0$, the quartic opens down and the range is $y \le c$.

Practice

1. (Warm-up) Describe the transformations of $y = x^3$ in $y = (x - 4)^3 + 2$, and give the point of symmetry.

Solution

Translation $4$ units right and $2$ units up. The point of symmetry is $(4, 2)$.

2. (Warm-up) Give the vertex and the range of $y = -(x + 1)^4 + 6$.

Solution

Vertex $(-1, 6)$. Since $a = -1 \lt 0$, the graph opens down: range $\{y \in \mathbb{R} \mid y \le 6\}$.

3. (Warm-up) Under the transformation $y = 3(x + 5)^3 - 1$, what is the image of the point $(2, 8)$ on $y = x^3$?

Solution

The mapping rule is $(x, y) \to (x - 5,\ 3y - 1)$:

$$(2, 8) \to (2 - 5,\ 3(8) - 1) = (-3, 23)$$

4. (Core) For $y = \tfrac{1}{2}(x + 2)^4 - 3$, describe the transformations, map the five key points of $y = x^4$, and state the range and the $y$-intercept.

Solution

Vertical compression by a factor of $\tfrac{1}{2}$, translation $2$ left and $3$ down. Mapping rule: $(x, y) \to \left(x - 2,\ \tfrac{1}{2}y - 3\right)$.

$y = x^4$	$(-2, 16)$	$(-1, 1)$	$(0, 0)$	$(1, 1)$	$(2, 16)$
image	$(-4, 5)$	$(-3, -2.5)$	$(-2, -3)$	$(-1, -2.5)$	$(0, 5)$

Range $\{y \in \mathbb{R} \mid y \ge -3\}$. The $y$-intercept is $5$ (it’s the image of $(2, 16)$; check: $\tfrac{1}{2}(2)^4 - 3 = 8 - 3 = 5$ ✓).

5. (Core) For $y = -(3x - 6)^3$, find $a$, $k$, $d$, and $c$, and map the points $(-1, -1)$, $(0, 0)$, and $(1, 1)$.

Solution

Factor: $y = -\big(3(x - 2)\big)^3$. So $a = -1$, $k = 3$, $d = 2$, $c = 0$.

Mapping rule: $(x, y) \to \left(\tfrac{x}{3} + 2,\ -y\right)$.

$$(-1, -1) \to \left(\tfrac{5}{3}, 1\right), \quad (0, 0) \to (2, 0), \quad (1, 1) \to \left(\tfrac{7}{3}, -1\right)$$

Check: at $x = \tfrac{7}{3}$, $-(7 - 6)^3 = -1$. ✓

6. (Core) A quartic of the form $y = a(x - d)^4 + c$ has vertex $(-3, -2)$ and passes through $(-1, 30)$. Find its equation.

Solution

$y = a(x + 3)^4 - 2$. Substitute $(-1, 30)$:

$$30 = a(2)^4 - 2 \quad\Rightarrow\quad 16a = 32 \quad\Rightarrow\quad a = 2$$

The equation is $y = 2(x + 3)^4 - 2$.

7. (Core) Explain why reflecting $y = x^3$ in the $y$-axis gives the same graph as reflecting it in the $x$-axis, but reflecting $y = x^4$ in the $y$-axis changes nothing.

Solution

A reflection in the $y$-axis replaces $x$ with $-x$.

For $y = x^3$: $(-x)^3 = -x^3$, which is the reflection in the $x$-axis.

For $y = x^4$: $(-x)^4 = x^4$, so the graph is unchanged. It’s already symmetric about the $y$-axis. (This is the idea of odd and even functions.)

8. (Challenge) Find the $x$-intercepts of $y = 2(x - 1)^4 - 32$, and state the range.

Solution

$$2(x - 1)^4 = 32 \quad\Rightarrow\quad (x - 1)^4 = 16 \quad\Rightarrow\quad x - 1 = \pm 2$$

So $x = 3$ or $x = -1$. (A fourth power is never negative, so $x - 1 = \pm 2$ are the only real solutions.)

Vertex $(1, -32)$ with $a \gt 0$: range $\{y \in \mathbb{R} \mid y \ge -32\}$.

9. (Challenge) For $y = -\tfrac{1}{4}(2x + 4)^3 + 2$, describe all the transformations of $y = x^3$, and find the point of symmetry, the $x$-intercept, and the $y$-intercept.

Solution

Factor: $y = -\tfrac{1}{4}\big(2(x + 2)\big)^3 + 2$. So $a = -\tfrac{1}{4}$, $k = 2$, $d = -2$, $c = 2$:

• vertical compression by a factor of $\tfrac{1}{4}$ and a reflection in the $x$-axis;
• horizontal compression by a factor of $\tfrac{1}{2}$;
• translation $2$ left and $2$ up.

Point of symmetry: $(-2, 2)$.

$x$-intercept:

$$-\tfrac{1}{4}(2x + 4)^3 + 2 = 0 \quad\Rightarrow\quad (2x + 4)^3 = 8 \quad\Rightarrow\quad 2x + 4 = 2 \quad\Rightarrow\quad x = -1$$

$y$-intercept: $-\tfrac{1}{4}(4)^3 + 2 = -16 + 2 = -14$.