Let ​f(x)=x2+10x+37​ . What is the vertex form off(x)?

What is the minimum value off(x)?

Enter your answers in the boxes.

Vertex form: f(x)=

Minimum value of f(x):

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

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

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

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