Describe the transformation that takes y = x² to y = (x − 3)² + 2, then find the vertex of the new graph.
Solution The term (x − 3) is a horizontal translation 3 units to the right, and the + 2 is a vertical translation 2 units up. The new vertex is (3, 2).
Example 2: an absolute value function
Describe the graph of y = |x + 1| − 4 and find its vertex.
Solution The (x + 1) gives a translation 1 unit to the left, and the − 4 gives a translation 4 units down. The vertex is (−1, −4), and the graph has a V-shape opening upward from this point.
Example 3: a reflection with a translation
Sketch y = −√x + 3
marking points
and state its domain and range.
Solution Start with y = √x, reflect it in the x-axis to get −√x, then move it 3 units up. At x = 0, y = 3. At x = 1, y = 2. At x = 4, y = 1.
The domain is [0, ∞[ and the range is ]−∞, 3].
Example 4: a stretch with a translation
The function f(x) = x³ is transformed into g(x) = 2(x − 1)³ − 5.
Describe the transformation and find the image of the point (0, 0).
Solution The transformation is a translation 1 unit to the right, then a vertical stretch by factor 2, then a translation 5 units down. The point (0, 0) moves to x = 1, and its y-value becomes 2 × 0 − 5 = −5. So the image of (0, 0) is (1, −5).
Check g(1) = 2 × 0 − 5 = −5, which agrees with the result.
Example 5: the image of a point after several transformations
The point (2, 6) lies on the graph of f. Find the image of this point on the graph of y = −f(x + 1).
Solution The translation 1 unit to the left changes x = 2 into x = 1. The reflection in the x-axis changes y = 6 into y = −6. So the image is (1, −6).
Example 6: writing the equation after a transformation
The graph of y = x² is moved 2 units to the left and 5 units down. Write the equation of the new graph and find its vertex.
Solution The move 2 units to the left adds 2 inside the bracket: y = (x + 2)². Then the move 5 units down gives y = (x + 2)² − 5. The new vertex is (−2, −5).
Graphing Functions · 3 / 5
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