Question

Let ​f(x)=x2+10x+37​ .

Answer 1

Answer:

f(x) = (x+5)^2 +12

The minimum value is 12

Step-by-step explanation:

f(x)=x^2+10x+37

The vertex will be the minimum value since this is an upwards opening parabola

Completing the square by taking the coefficient of x and squaring it adding it and subtracting it

f(x) = x^2+10x  + (10/2) ^2 - (10/2) ^2+37

f(x) = ( x^2 +10x +25) -25+37

    = ( x+5) ^2+12

Th is in vertex form y = ( x-h)^2 +k  where (h,k) is the vertex

The vertex is (-5,12)

The minimum is the y value or 12