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Algebra · real student question

The function f(x) = |x + 2| - 3 is reflected in the x-axis and then translated 2 units down. Write the equation of the new function.

Question

The function f(x)=∣x+2∣−3f(x)=|x+2|-3 is reflected in the xx-axis and then translated 22 units down. Write the equation of the resulting function.

Step-by-step solution

  1. Read the starting vertex. f(x)=∣x+2∣−3f(x)=|x+2|-3 opens upward with vertex (−2,−3)(-2,-3), and it crosses the xx-axis twice, where ∣x+2∣=3|x+2|=3, that is at x=−5x=-5 and x=1x=1.

  2. Reflect in the xx-axis: negate everything. Every output changes sign, so the whole right-hand side gets a minus sign in front:

    −f(x)=−(∣x+2∣−3)=−∣x+2∣+3-f(x)=-\left(|x+2|-3\right)=-|x+2|+3

    The constant flips from −3-3 to +3+3. Distributing that minus over both terms is the entire difficulty of the problem.

  3. Translate 2 units down. Subtract 22 from the reflected function:

    −∣x+2∣+3−2=−∣x+2∣+1-|x+2|+3-2=-|x+2|+1

  4. Final equation and its shape.

    f(x)=−∣x+2∣+1,vertex (−2,1), opens downwardf(x)=-|x+2|+1,\qquad \text{vertex }(-2,1),\ \text{opens downward}

    The vertex went from (−2,−3)(-2,-3) to (−2,3)(-2,3) under reflection, then down 22 to (−2,1)(-2,1).

  5. Verify with points and intercepts. Original (0,−1)(0,-1) reflects to (0,1)(0,1) then shifts to (0,−1)(0,-1); the formula gives −∣0+2∣+1=−2+1=−1  ✓-|0+2|+1=-2+1=-1\;\checkmark. The new xx-intercepts satisfy ∣x+2∣=1|x+2|=1, so x=−3x=-3 and x=−1x=-1 — the two roots have moved inward from {−5,1}\{-5,1\}, consistent with a downward-opening V whose peak is only 11 unit above the axis.

Answer

f(x)=−∣x+2∣+1f(x)=-|x+2|+1

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