The function is reflected in the -axis and then translated units down. Write the equation of the resulting function.
Identify the starting graph. is a V opening upward with vertex at , since puts and the trailing puts .
Reflect in the -axis by negating the whole output. This is the step that trips people: the minus sign must apply to the entire function, constant included, not just the absolute-value term:
Writing at this stage would be the classic error — the becomes under reflection, because the point flips to .
Translate 2 units down by subtracting 2.
State the result and its features.
The vertex has returned to its original coordinates by coincidence — the reflection lifted it from to and the shift dropped it back by — but the graph is genuinely different, now an upside-down V.
Check with points. Original ; reflecting gives and shifting down gives . The final formula: . Original vertex , and the formula gives . Since the maximum value is , the graph never reaches the -axis.
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