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Home > CCA > Chapter 10 > Lesson 10.3.2 > Problem 10-125

10-125.
  1. Joaquin needed to find the maximum value of y = −x2 + 4x − 1 but was stuck. Homework Help ✎

  2. “No problem,”said Thao. “The parabola opens downward so just complete the square to find the vertex. The y-value of the vertex will be the maximum.”

  3. “I know that,” said Joaquin. “The problem is that it is −x2.”

  4. Mmm…what if we multiply everything by –1? Can we now complete the square?”

    1. Multiply the equation by –1 so you have −y = ....

    2. Complete the square on the right side.

    3. Multiply the equation by –1 so you are back to y = ...

    4. Identify the vertex and the maximum value.

Each term should have the opposite sign.

y = (x − 2)2 − 3

y = −(x − 2)2 + 3

If y = a(xh)2 + k, then the vertex is at (h, k).