### Home > CCAA8 > Chapter 10 Unit 11 > Lesson CCA: 10.3.2 > Problem10-125

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?”

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

Each term should have the opposite sign.

1. Complete the square on the right side.

$−y=(x−2)^2−3$

1. Multiply the equation by $-1$ so you are back to $y=$ ...

$y=−(x−2)^2+3$

1. Identify the vertex and the maximum value.

If $y=a(x−h)^2+k$, then the vertex is at $(h,k)$.