CPM Homework Help : CCA Problem 10-125

Joaquin needed to find the maximum value of $y=−x^2+4x−1$ but was stuck.

“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.”

“I know that,” said Joaquin. “The problem is that it is $−x^2$ .”

“Mmm…what if we multiply everything by $-1$ ? Can we now complete the square?”

Multiply the equation by $-1$ so you have $−y=$ ....
Hint (a):
Each term should have the opposite sign.

Complete the square on the right side.
Answer (b):
$−y=(x−2)^2−3$

Multiply the equation by $-1$ so you are back to $y=$ ...
Answer (c):
$y=−(x−2)^2+3$

Identify the vertex and the maximum value.
Hint (d):
If $y=a(x−h)^2+k$ , then the vertex is at $(h,k)$ .