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Functions - Symmetries of f(x) – more transformations

Grade 12IB_AA

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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An even function f(x)f(x) exhibits symmetry about the yy-axis, satisfying f(−x)=f(x)f(-x) = f(x). This means the point (x,y)(x, y) on the graph is reflected to (−x,y)(-x, y). Common examples include y=x2y = x^2 and y=cos⁡(x)y = \cos(x).

Graph of y = x squared showing symmetry across the y-axis.
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An odd function f(x)f(x) exhibits rotational symmetry of 180∘180^\circ about the origin, satisfying f(−x)=−f(x)f(-x) = -f(x). This means the point (x,y)(x, y) is reflected through the origin to (−x,−y)(-x, -y). Examples include y=x3y = x^3 and y=sin⁡(x)y = \sin(x).

Graph of y = x cubed showing rotational symmetry about the origin.
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The transformation y=∣f(x)∣y = |f(x)| reflects any portion of the graph where f(x)<0f(x) < 0 (below the xx-axis) across the xx-axis to the positive yy region, making the entire range non-negative.

Graph showing the absolute value transformation reflecting negative parts upwards.
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The transformation y=f(∣x∣)y = f(|x|) discards the portion of the graph where x<0x < 0 and replaces it with a reflection of the portion where x≥0x \geq 0 across the yy-axis. This ensures the resulting function is always even.

Graph of f(|x|) showing right-side reflection onto the left side.

📐Formulae

f(−x)=f(x) (Even Function)f(-x) = f(x) \text{ (Even Function)}

f(−x)=−f(x) (Odd Function)f(-x) = -f(x) \text{ (Odd Function)}

∣f(x)∣={f(x)if f(x)≥0−f(x)if f(x)<0|f(x)| = \begin{cases} f(x) & \text{if } f(x) \geq 0 \\ -f(x) & \text{if } f(x) < 0 \end{cases}

f(∣x∣)={f(x)if x≥0f(−x)if x<0f(|x|) = \begin{cases} f(x) & \text{if } x \geq 0 \\ f(-x) & \text{if } x < 0 \end{cases}

y=a⋅f(b(x−h))+k (General Transformation Form)y = a \cdot f(b(x - h)) + k \text{ (General Transformation Form)}

💡Examples

Problem 1:

Determine algebraically whether the function f(x)=x3x2+1f(x) = \frac{x^3}{x^2 + 1} is even, odd, or neither.

Solution:

To test for symmetry, we find f(−x)f(-x): f(−x)=(−x)3(−x)2+1f(-x) = \frac{(-x)^3}{(-x)^2 + 1} f(−x)=−x3x2+1f(-x) = \frac{-x^3}{x^2 + 1} Since f(−x)=−(x3x2+1)f(-x) = -\left(\frac{x^3}{x^2 + 1}\right), we have f(−x)=−f(x)f(-x) = -f(x).

Explanation:

Because f(−x)=−f(x)f(-x) = -f(x), the function satisfies the condition for an odd function. It possesses rotational symmetry about the origin.

Problem 2:

Given the function g(x)=x2−4x+3g(x) = x^2 - 4x + 3, describe the transformation required to obtain the graph of h(x)=2g(x+1)−5h(x) = 2g(x + 1) - 5.

Solution:

  1. Horizontal translation: Shift left by 11 unit (x→x+1x \to x + 1).
  2. Vertical stretch: Stretch vertically by a scale factor of 22 (g→2gg \to 2g).
  3. Vertical translation: Shift down by 55 units (2g→2g−52g \to 2g - 5).

Explanation:

Transformations inside the function argument affect xx (horizontal), while transformations outside affect yy (vertical). Following the order of operations, we handle the horizontal shift, then the vertical stretch, and finally the vertical shift.

Problem 3:

Let f(x)=x−2f(x) = x - 2. Sketch the graph of y=f(∣x∣)y = f(|x|) and state its yy-intercept.

Solution:

For x≥0x \geq 0, f(∣x∣)=f(x)=x−2f(|x|) = f(x) = x - 2. For x<0x < 0, we reflect the part where x≥0x \geq 0 across the yy-axis, resulting in f(−x)=−x−2f(-x) = -x - 2. The yy-intercept is found at f(∣0∣)f(|0|): y=0−2=−2y = 0 - 2 = -2 So the yy-intercept is (0,−2)(0, -2).

Explanation:

The transformation f(∣x∣)f(|x|) creates a mirror image of the right side of the graph onto the left side, making the resulting function even.

Problem 4:

Sketch the graph of y=∣2x−4∣y = |2x - 4| and find the coordinates of its vertex and yy-intercept.

V-shaped graph of y = |2x - 4| with vertex at (2,0).

Solution:

  1. Start with the linear function f(x)=2x−4f(x) = 2x - 4. Its xx-intercept is at 2x−4=0  ⟹  x=22x - 4 = 0 \implies x = 2. Its yy-intercept is at (0,−4)(0, -4).
  2. Apply the absolute value transformation y=∣f(x)∣y = |f(x)|. The portion of the line below the xx-axis (where x<2x < 2) is reflected upwards.
  3. The vertex of the V-shape is the former xx-intercept at (2,0)(2, 0).
  4. The new yy-intercept is ∣−4∣=4|-4| = 4, resulting in the point (0,4)(0, 4).

Explanation:

The modulus transformation y=∣f(x)∣y = |f(x)| acts as a reflection in the xx-axis for all negative outputs of the function.

Problem 5:

Given f(x)=(x−2)2f(x) = (x-2)^2, sketch the graph of g(x)=f(∣x∣)g(x) = f(|x|).

Graph of y = (|x|-2)^2 showing two symmetric parabolic dips.

Solution:

  1. The original graph f(x)=(x−2)2f(x) = (x-2)^2 is a parabola with a vertex at (2,0)(2, 0).
  2. To graph f(∣x∣)f(|x|), we keep the portion of the graph where x≥0x \geq 0. This includes the vertex at (2,0)(2, 0) and the yy-intercept at (0,4)(0, 4).
  3. We ignore the original graph for x<0x < 0 and instead reflect the x≥0x \geq 0 portion across the yy-axis.
  4. This results in a 'W-shaped' curve with vertices at (2,0)(2, 0) and (−2,0)(-2, 0).

Explanation:

The f(∣x∣)f(|x|) transformation creates an even function by mirroring the right side of the graph onto the left.