To write f(x) = 2x2 – 44x + 185, factor out from the first two terms. next, form a perfect square trinomial keeping the value of the function equivalent: f(x) = 2(x2 – 22x + 121) + 185 – 242 the function written in vertex form is f(x) = (x – )2 + .

f(x) = 2(x-11)^2 - 57

Step-by-step explanation:

To write f(x) = 2x^2 – 44x + 185, factor out from the first two terms.

Next, form a perfect square tri nomial keeping the value of the function equivalent: f(x) = 2(x^2 – 22x + 121) + 185 – 242

To get vertex form we factor x^2 -22x+121

product is 121 and sum is -22

-11*-11= 121 and sum -11+(-11)= -22

x^2 -22x+121

(x-11)(x-11)

f(x) = 2(x^2 – 22x + 121) + 185 – 242

f(x) = 2(x-11)(x-11)+ 185 – 242

f(x) = 2(x-11)^2 - 57

tell you what. i'll find the length of the medians and you can fill in the blanks where they belong.

br

b = (2a,0)

r = (a/2,b/2) remember these things are medians. they go to the 1/2 way point of the line opposite the vertex they face.

br^2 =   (2a - a/2)^2 + (0 - b/2)^2

br^2 = (3/2 a) ^2 + b^2 / 4

br^2 = 9/4 a^2 + b^2 / 4   we need to find some relationship between a and b.

let's try ab = bc  

ab = 2a

bc = (2a - a)^2 + (b - 0)^2

bc = sqrt(a^2 + b^2)

ab = bc

2a = sqrt(a^2 + b^2) square both sides.

4a^2 = a^2 + b^2 subtract a^2 from both sides.

sqrt(3a^2) = sqrt(b^2)

sqrt(3)a = b

let's put b into br^2

br^2 = 9/4 a^2 + 3b^2 / 4

br^2 = 12 a^2 / 4

br^2 = 3a^2 

br = sqrt(3) * a

cp

c = (a,b) ; p = (a,0) this is another application of the distance formula, and it is a good one.

cp^2 = (a - a)^2 + (b - 0)^2

cp^2 = 0 + b^2

cp^2 = b^2 which you can see from the diagram. 

cp = sqrt(b^2)

cp = b but b = sqrt(3) * a 

cp = sqrt(3)*a

aq

a = (0,0)

q = (3/2) a, b/2

aq^2 = (3a/2 -0)^2 + (b/2 - 0)^2

aq^2 = 9a^2/4 + b^2/4

but b = sqrt(3) * a 

aq^2 = 9a^2 / 4 + (sqrt(3)*a)^2 /4

aq^2 = 9 a^2/ 4 + 3 a^2 / 4

aq^2 = 12 a^2/4

aq^2 = 3 a^2

aq = (sqrt  3) * a which agrees with the other two.