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Functions - Function transformations

Grade 12IB_AI

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

🔑Concepts

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Vertical and Horizontal Translations: A vertical translation moves the graph up or down, represented by y=f(x)+ky = f(x) + k. A horizontal translation moves the graph left or right, represented by y=f(x−h)y = f(x - h). Note that a positive hh moves the graph to the right, while a negative hh moves it to the left.

Graph showing the transformation of f(x) = x^2 to f(x-2)+1, illustrating a shift right by 2 and up by 1.
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Reflections: A reflection in the xx-axis occurs when the entire function is multiplied by −1-1, expressed as y=−f(x)y = -f(x). A reflection in the yy-axis occurs when the input variable xx is replaced by −x-x, expressed as y=f(−x)y = f(-x).

Graph showing an exponential function reflected across the x-axis.
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Vertical Stretching and Compression: Multiplying the function by a constant aa results in a vertical stretch by factor aa if ∣a∣>1|a| > 1, or a vertical compression if 0<∣a∣<10 < |a| < 1. All yy-coordinates are multiplied by aa, while xx-coordinates remain unchanged.

Graph of absolute value function stretched vertically by a factor of 2.
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Horizontal Stretching and Compression: Multiplying the input xx by a constant bb results in a horizontal stretch by factor 1b\frac{1}{b} if 0<∣b∣<10 < |b| < 1, or a horizontal compression if ∣b∣>1|b| > 1. This transformation affects the xx-intercepts and widths of features.

Sine wave showing horizontal compression when the input is multiplied by 2.

📐Formulae

y=f(x)+k (Vertical Translation)y = f(x) + k \text{ (Vertical Translation)}

y=f(x−h) (Horizontal Translation)y = f(x - h) \text{ (Horizontal Translation)}

y=a⋅f(x) (Vertical Stretch by factor a)y = a \cdot f(x) \text{ (Vertical Stretch by factor } a)

y=f(b⋅x) (Horizontal Stretch by factor 1b)y = f(b \cdot x) \text{ (Horizontal Stretch by factor } \frac{1}{b})

y=−f(x) (Reflection in x-axis)y = -f(x) \text{ (Reflection in } x\text{-axis)}

y=f(−x) (Reflection in y-axis)y = f(-x) \text{ (Reflection in } y\text{-axis)}

💡Examples

Problem 1:

Given the function f(x)=x2f(x) = x^2, describe the transformations required to obtain the graph of g(x)=3(x+2)2−5g(x) = 3(x + 2)^2 - 5.

Solution:

  1. Horizontal translation left by 22 units (h=−2h = -2).
  2. Vertical stretch by a scale factor of 33 (a=3a = 3).
  3. Vertical translation down by 55 units (k=−5k = -5).

Explanation:

In the form a⋅f(x−h)+ka \cdot f(x - h) + k, we identify h=−2h = -2 (since it is x−(−2)x - (-2)), a=3a = 3, and k=−5k = -5.

Problem 2:

A point P(4,10)P(4, 10) lies on the graph of y=f(x)y = f(x). Find the coordinates of the corresponding point P′P' on the graph of y=f(2x)+1y = f(2x) + 1.

Solution:

x′=42=2x' = \frac{4}{2} = 2 y′=10+1=11y' = 10 + 1 = 11 Therefore, P′=(2,11)P' = (2, 11).

Explanation:

The transformation f(2x)f(2x) is a horizontal stretch with scale factor 12\frac{1}{2}, so we divide the xx-coordinate by 22. The +1+1 is a vertical translation, so we add 11 to the yy-coordinate.

Problem 3:

The function f(x)=xf(x) = \sqrt{x} is reflected in the xx-axis and then translated 44 units to the right. Write the equation of the resulting function h(x)h(x).

Solution:

h(x)=−x−4h(x) = -\sqrt{x - 4}

Explanation:

Reflection in the xx-axis changes f(x)f(x) to −f(x)=−x-f(x) = -\sqrt{x}. Translation 44 units right replaces xx with (x−4)(x - 4), resulting in −x−4-\sqrt{x - 4}.

Problem 4:

The graph of y=f(x)y = f(x) is shown below as a semi-circle centered at the origin with radius 22. Sketch the graph of g(x)=f(x−3)−2g(x) = f(x - 3) - 2 and state the new center of the semi-circle.

Graph showing a semi-circle translated 3 units right and 2 units down.

Solution:

  1. Identify the transformations: f(x−3)f(x - 3) is a horizontal translation 33 units to the right. −2-2 is a vertical translation 22 units down.
  2. The original center is at (0,0)(0, 0). Applying the translations, the new center becomes (0+3,0−2)=(3,−2)(0 + 3, 0 - 2) = (3, -2).
  3. The radius remains 22. The domain of the original function was [−2,2][-2, 2]; the new domain is [1,5][1, 5]. The range of the original was [0,2][0, 2]; the new range is [−2,0][-2, 0].

Explanation:

To transform the graph, move every point on the original semi-circle 33 units to the right and 22 units down.

Problem 5:

Consider the function f(x)=1xf(x) = \frac{1}{x}. If g(x)=f(−x)+3g(x) = f(-x) + 3, describe the sequence of transformations and sketch the resulting graph including the horizontal asymptote.

Graph of y = -1/x + 3 showing a horizontal asymptote at y=3.

Solution:

  1. The transformation f(−x)f(-x) is a reflection in the yy-axis.
  2. The transformation +3+ 3 is a vertical translation 33 units up.
  3. Original horizontal asymptote: y=0y = 0. New horizontal asymptote: y=3y = 3.
  4. Original vertical asymptote: x=0x = 0. Reflection in yy-axis leaves this unchanged at x=0x = 0.

Explanation:

Applying a reflection in the yy-axis to 1x\frac{1}{x} results in −1x-\frac{1}{x}. Moving this up by 3 units shifts the entire structure, including the horizontal asymptote, to y=3y=3.