Compete the square to determine minum or maxuim value of function define by -x2+10x+5

maximum value y = 30

Step-by-step explanation:

Given

- x² + 10x + 5

To complete the square the coefficient of the x² term must be 1

factor out - 1

= - (x² - 10x) + 5

To complete the square

add/subtract ( half the coefficient of the x- term )² to x² - 10x

= - (x² + 2(- 5)x + 25 - 25) + 5

= - (x - 5)² + 25 + 5

= - (x - 5)² + 30 ← in vertex form

y = a(x - h)² + k

where (h, k) are the coordinates of the vertex and a is a multiplier

Hence vertex = (5, 30)

The max/ min occurs at the vertex

Since a < 0 then vertex is a maximum

Hence maximum value is y = 30