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Of Graph Dse Exercise | Transformation

Before we dive into the exercise, ensure you have this table memorized. Let the equation of the graph be $y = f(x)$.

| Transformation | New Equation | Effect on Graph | | :--- | :--- | :--- | | Vertical Translation | $y = f(x) + k$ | Shift up by $k$ units (if $k > 0$). | | | $y = f(x) - k$ | Shift down by $k$ units. | | Horizontal Translation | $y = f(x - k)$ | Shift right by $k$ units. | | | $y = f(x + k)$ | Shift left by $k$ units. | | Reflection | $y = -f(x)$ | Reflect about the x-axis. | | | $y = f(-x)$ | Reflect about the y-axis. | | Scaling (Stretch/Compress) | $y = k \cdot f(x)$ | Vertical stretch by factor $k$ (if $k > 1$). | | | $y = f(kx)$ | Horizontal compression by factor $\frac1k$. |

⚠️ The DSE Trap: The most common mistake in DSE exams is horizontal translation and scaling.

Question: The graph of $y = f(x)$ is transformed such that the x-coordinates are halved, and the y-coordinates are doubled. Which of the following represents the new equation? transformation of graph dse exercise

A. $y = 2f(2x)$ B. $y = \frac12f(2x)$ C. $y = 2f(\frac12x)$ D. $y = \frac12f(\frac12x)$

👀 Think about it first!

. . .

Solution: This is a classic DSE trap.

  • y-coordinates are doubled: The graph is stretched vertically.

    • Combining them: $y = 2f(2x)$.

    Correct Answer: A

    Answers:

  • Horizontal Stretch/Compression: ( y = f(bx) )

    • Start from ( f(x) = \sqrtx ).

  • Outside: Multiply by ( -2 ).
  • Final graph: Domain ( (-\infty, 4] ). Starts at ( (4, 0) ), then decreases sharply leftwards.

    • Answers:

    Given (y = \sqrtx) for (x \ge 0), sketch (y = -\sqrtx+2 + 1).

    Solution: