C. Polynomial and Rational Functions · Polynomial Functions and Equations

Transformations of y = x³ and y = x⁴

C1.6About 70 minutes

Learning goals

I can describe the roles of a, k, d and c in y = a[k(x − d)]ⁿ + c for n = 3 and n = 4.
I can sketch a transformed cubic or quartic by mapping key points with (x, y) → (x/k + d, ay + c).
I can write the equation of a transformed cubic or quartic from a description or from a graph.
I can find intercepts, the range, and the centre or vertex of a transformed cubic or quartic.

Words to know

parent function
The simplest function of a family, such as y = x³ or y = x⁴, from which others are made by transformations.

mapping notation
A rule such as (x, y) → (x/k + d, ay + c) that says where each point of the parent graph moves.

vertex (of a quartic)
The turning point of y = a[k(x − d)]⁴ + c, at (d, c).

centre (of a cubic)
The point of symmetry of y = a[k(x − d)]³ + c, at (d, c). The graph flattens there; it is a point of inflection.

Same moves, new parents

Remember from Grade 11: in Functions and Equivalent Expressions you transformed y = f(x) into y = af[k(x − d)] + c:

a: vertical stretch by a factor |a| (compression if |a| < 1); reflection in the x-axis if a < 0.
k: horizontal stretch by a factor 1/|k| (so k = 2 is a compression by 1/2); reflection in the y-axis if k < 0.
d: horizontal translation (right if d > 0).
c: vertical translation (up if c > 0).

Apply stretches and reflections first, then translations. In mapping notation: (x, y) → (x/k + d, ay + c).

Now the parents are y = x³ and y = x⁴. Their key points are:

y = x³: (−2, −8), (−1, −1), (0, 0), (1, 1), (2, 8). The origin is the centre.
y = x⁴: (−2, 16), (−1, 1), (0, 0), (1, 1), (2, 16). The origin is the vertex.

After the transformation, the centre or vertex moves to (d, c).

Worked example

Example 1: A transformed cubic

Describe the transformations in y = −2(x − 3)³ + 1, and use key points to sketch it.

a = −2: reflect in the x-axis and stretch vertically by a factor of 2. k = 1. d = 3: shift right 3. c = 1: shift up 1.
Mapping: (x, y) → (x + 3, −2y + 1).
(−1, −1) → (2, 3); (0, 0) → (3, 1); (1, 1) → (4, −1); (−2, −8) → (1, 17); (2, 8) → (5, −15).
The centre is (3, 1). Since a < 0, the graph rises on the left and falls on the right (Q2 to Q4).

See the graph: the centre moves to (3, 1) and the curve is flipped and twice as steep.

y = x³ (dashed) and y = −2(x − 3)³ + 1 (solid). The points (−1, −1), (0, 0), (1, 1) map to (2, 3), (3, 1), (4, −1).

Worked example

Example 2: A transformed quartic with k

Sketch y = ½[2(x + 1)]⁴ − 3 using key points, and state the vertex and range.

a = ½: vertical compression by ½. k = 2: horizontal compression by ½. d = −1: shift left 1. c = −3: shift down 3.
Mapping: (x, y) → (x/2 − 1, y/2 − 3).
(−2, 16) → (−2, 5); (−1, 1) → (−1.5, −2.5); (0, 0) → (−1, −3); (1, 1) → (−0.5, −2.5); (2, 16) → (0, 5).
Check one point in the equation: x = 0 gives ½(2)⁴ − 3 = 8 − 3 = 5 ✓.
a > 0, so the graph opens up from its vertex (−1, −3).

Vertex (−1, −3), range y ≥ −3.

y = x⁴ (dashed) and y = ½[2(x + 1)]⁴ − 3 (solid). The vertex moves from (0, 0) to (−1, −3); the points (±1, 1) move to (−1.5, −2.5) and (−0.5, −2.5).

Two shortcuts for power functions

1. k can be traded for a. Since (kx)ⁿ = kⁿxⁿ, a horizontal compression by 1/k gives the same graph as a vertical stretch by kⁿ. For example, y = ½[2(x + 1)]⁴ − 3 is the same as y = 8(x + 1)⁴ − 3.

2. Reflections in the y-axis. (−x)⁴ = x⁴, so reflecting y = x⁴ in the y-axis changes nothing (it is even). (−x)³ = −x³, so for y = x³ a reflection in the y-axis gives the same graph as a reflection in the x-axis (it is odd).

Worked example

Example 3: Equations from descriptions and graphs

(a) y = x⁴ is reflected in the x-axis, stretched vertically by a factor of 3 and translated 2 left and 5 up. Write the equation and give the range. (b) A cubic of the form y = a(x − d)³ + c has its centre at (2, −3) and passes through (3, −1). Find its equation.

(a) a = −3, d = −2, c = 5: y = −3(x + 2)⁴ + 5. It opens down from the vertex (−2, 5), so the range is y ≤ 5.
(b) The centre gives d = 2 and c = −3: y = a(x − 2)³ − 3.
Substitute (3, −1): −1 = a(1)³ − 3, so a = 2.

(a) y = −3(x + 2)⁴ + 5, range y ≤ 5 (b) y = 2(x − 2)³ − 3

Worked example

Example 4: Intercepts of transformed functions

Find the x-intercepts of (a) y = (x + 1)⁴ − 16 (b) y = −2(x − 3)³ + 16, and the y-intercept of each.

(a) Set y = 0: (x + 1)⁴ = 16, so x + 1 = 2 or x + 1 = −2 (fourth roots come in ± pairs). x = 1 or x = −3. y-intercept: (1)⁴ − 16 = −15.
(b) Set y = 0: −2(x − 3)³ = −16, so (x − 3)³ = 8 and x − 3 = 2 (a cube root has only one real value). x = 5. y-intercept: −2(−3)³ + 16 = 54 + 16 = 70.

(a) x = 1, −3; y = −15 (b) x = 5; y = 70

Just for fun

The cubic y = x³ is its own upside-down twin: reflect it in the x-axis or reflect it in the y-axis and you get exactly the same new graph, y = −x³. Two different mirrors, one reflection.

Hands-on activity

Cardboard template transformations

You need: a piece of thin cardboard (from a cereal box), grid paper with 1 cm squares, scissors (use them carefully), a pencil and a ruler.

On grid paper (1 square = 1 unit), plot y = x⁴ through (−1, 1), (0, 0), (1, 1) and the points (±1.5, 5.06). Draw a smooth curve.
Trace the curve onto the cardboard, cut along it, and mark the vertex point on the template.
On a fresh grid, use the template to draw y = (x − 2)⁴ − 3 (slide the vertex to (2, −3)) and y = −(x + 1)⁴ + 4 (flip the template upside down, vertex at (−1, 4)).
Explain why you cannot use the same template for y = 2(x − 2)⁴ − 3.

What to look for: sliding and flipping the template handles c, d and the sign of a. A stretch (|a| = 2) changes the shape itself, so it needs a new template (or plotting points with the mapping rule).

Your turn

Describe the transformations in y = 4(x + 2)³ − 5.

Under y = −(x − 4)⁴ + 2, what is the image of the point (1, 1) on y = x⁴?

Write the equation: y = x⁴ compressed vertically by a factor of ½, then translated 3 right and 1 down.

Give the vertex and range of y = −3(x + 1)⁴ + 6.

Is y = (−x)³ the same graph as y = −x³?

Map (−1, −1), (0, 0), (1, 1) onto y = (x − 2)³ + 1.

Find the x-intercept of y = 2(x − 1)³ − 54.

More practice: warm-up

In y = a[k(x − d)]ⁿ + c, which parameter shifts the graph left or right?

What does a negative value of a do?

What is the vertex of y = (x + 5)⁴?

How is y = x³ + 7 related to y = x³?

Does y = −x⁴ open up or down?

More practice: on your own

Describe y = [3(x − 1)]³ + 2 and write it with k = 1. Check one mapped point.

A quartic y = a(x − d)⁴ + c has vertex (−2, 1) and passes through (0, −15). Find a.

A cubic y = a(x − d)³ + c has centre (1, 4) and passes through (−1, 0). Find its equation.

Find the x-intercepts of y = −(x − 2)⁴ + 81.

Simplify y = −½(−x + 3)³ to the form y = a(x − d)³ and describe it.

Challenge

Explain without a graph why y = 2(x − 1)⁴ + 3 has no x-intercepts.

Find the transformation y = a(x − d)⁴ + c of y = x⁴ whose graph has x-intercepts 0 and 2 and a minimum value of −4.

Show that y = (2x)³ and y = 8x³ are the same function, and find the vertical stretch that matches y = (2x)⁴.

Check your understanding

Quiz

What is the vertex of y = −2(x − 4)⁴ + 7?

Which equation is y = x³ stretched vertically by 3, translated 1 left and 2 down?

Under y = ½(x + 2)³ − 1, the point (2, 8) on y = x³ maps to which point?

What is the range of y = −(x + 3)⁴ − 2?

Which function has the same graph as y = (3x)⁴?

What are the x-intercepts of y = (x − 1)⁴ − 16?

For parents and teachers

Big idea (C1.6): the transformation rules from Grade 11 work for any parent function. For y = a[k(x − d)]ⁿ + c, the centre (cubic) or vertex (quartic) moves to (d, c), and each point maps by (x, y) → (x/k + d, ay + c).

How to help: Ask your teen to read an equation aloud as a list of moves ("reflect, stretch by 2, right 3, up 1") and to show where (1, 1) goes. Graphing it in Desmos afterwards is a quick check.

Common difficulties:

Moving the wrong way: (x − 3) shifts right, (x + 3) shifts left.
Treating k = 2 as a stretch by 2 (it is a compression by ½).
Applying translations before stretches when mapping points.
Forgetting the ± when taking a fourth root, or adding a ± to a cube root (a cube root has only one real value).