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Functions - Transformations of functions

Grade 11IB_AA

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

🔑Concepts

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Vertical translations y=f(x)+ky = f(x) + k shift the graph up (k>0k > 0) or down (k<0k < 0), while horizontal translations y=f(x−h)y = f(x - h) shift the graph right (h>0h > 0) or left (h<0h < 0).

Graph showing the transformation of f(x) = x^2 to f(x-2)+3, shifting right 2 units and up 3 units.
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Vertical stretching y=a⋅f(x)y = a \cdot f(x) pulls the graph away from the xx-axis by scale factor aa, whereas vertical compression occurs if 0<a<10 < a < 1. Reflection in the xx-axis is given by y=−f(x)y = -f(x).

Comparison of cos(x), 2*cos(x) (vertical stretch), and -cos(x) (reflection in x-axis).
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Horizontal stretching/compression y=f(qx)y = f(qx) uses a scale factor of 1q\frac{1}{q}. If q>1q > 1, the graph is compressed towards the yy-axis; if 0<q<10 < q < 1, it is stretched away from the yy-axis.

Horizontal compression of the square root function.
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Reflection in the yy-axis is achieved by replacing xx with −x-x, resulting in the function y=f(−x)y = f(-x). This maps points (x,y)(x, y) to (−x,y)(-x, y).

📐Formulae

y=f(x)+k(Vertical translation by k units)y = f(x) + k \quad \text{(Vertical translation by } k \text{ units)}

y=f(x−h)(Horizontal translation by h units)y = f(x - h) \quad \text{(Horizontal translation by } h \text{ units)}

y=a⋅f(x)(Vertical stretch, scale factor a)y = a \cdot f(x) \quad \text{(Vertical stretch, scale factor } a\text{)}

y=f(qx)(Horizontal stretch, scale factor 1q)y = f(qx) \quad \text{(Horizontal stretch, scale factor } \frac{1}{q}\text{)}

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

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

💡Examples

Problem 1:

The function f(x)=x2f(x) = x^2 is translated by the vector (3−2)\begin{pmatrix} 3 \\ -2 \end{pmatrix} and then reflected in the xx-axis. Find the equation of the resulting function g(x)g(x).

Solution:

g(x)=−((x−3)2−2)=−(x−3)2+2g(x) = -((x - 3)^2 - 2) = -(x - 3)^2 + 2

Explanation:

Step 1: Apply the translation (3−2)\begin{pmatrix} 3 \\ -2 \end{pmatrix} to f(x)f(x), which gives f(x−3)−2f(x - 3) - 2, resulting in (x−3)2−2(x - 3)^2 - 2. Step 2: To reflect in the xx-axis, multiply the entire function by −1-1. This gives −((x−3)2−2)-( (x - 3)^2 - 2 ), which simplifies to −(x−3)2+2-(x - 3)^2 + 2.

Problem 2:

Describe the transformations required to transform the graph of f(x)=ln⁡(x)f(x) = \ln(x) into the graph of g(x)=3ln⁡(2x)g(x) = 3\ln(2x).

Solution:

  1. A horizontal stretch with a scale factor of 12\frac{1}{2}. 2. A vertical stretch with a scale factor of 33.

Explanation:

The term 2x2x inside the function represents a horizontal stretch by a factor of 1q=12\frac{1}{q} = \frac{1}{2}. The multiplier 33 outside the function represents a vertical stretch by a scale factor of a=3a = 3.

Problem 3:

The 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=2f(x−1)+5y = 2f(x - 1) + 5.

Solution:

P′(5,25)P'(5, 25)

Explanation:

First, handle the horizontal change: x−1x - 1 means we add 11 to the xx-coordinate, so 4+1=54 + 1 = 5. Next, handle the vertical changes: the yy-coordinate is multiplied by 22 and then 55 is added. So, (10×2)+5=25(10 \times 2) + 5 = 25. The new point is (5,25)(5, 25).

Problem 4:

The graph of f(x)=∣x∣f(x) = |x| is transformed into the graph of g(x)=−∣x+2∣+4g(x) = -|x + 2| + 4. Sketch the graph of g(x)g(x) and identify the coordinates of its vertex.

Graph of g(x) = -|x+2|+4 showing an inverted V-shape with vertex at (-2, 4).

Solution:

  1. Start with f(x)=∣x∣f(x) = |x|, which has a vertex at (0,0)(0, 0).
  2. Shift left by 22 units: f(x+2)=∣x+2∣f(x + 2) = |x + 2|, vertex at (−2,0)(-2, 0).
  3. Reflect in the xx-axis: −f(x+2)=−∣x+2∣-f(x + 2) = -|x + 2|, vertex remains at (−2,0)(-2, 0).
  4. Shift up by 44 units: g(x)=−∣x+2∣+4g(x) = -|x + 2| + 4, vertex at (−2,4)(-2, 4). The resulting vertex is (−2,4)(-2, 4).

Explanation:

Multiple transformations are applied in sequence: horizontal translation, reflection, then vertical translation.

Problem 5:

Given f(x)=exf(x) = e^x, find the new function h(x)h(x) after a horizontal stretch with scale factor 33 followed by a reflection in the yy-axis.

Graph showing the reflection of e^(x/3) in the y-axis to produce e^(-x/3).

Solution:

  1. Horizontal stretch by scale factor 33 means q=13q = \frac{1}{3}. The function becomes f(13x)=e13xf(\frac{1}{3}x) = e^{\frac{1}{3}x}.
  2. Reflection in the yy-axis replaces xx with −x-x. The function becomes h(x)=e13(−x)=e−x3h(x) = e^{\frac{1}{3}(-x)} = e^{-\frac{x}{3}}.

Explanation:

The horizontal stretch uses the reciprocal of the scale factor inside the function argument, and the reflection negates the xx variable.

Transformations of functions Grade 11 Notes & Examples