What is the range of the function f(x) = 3x2 + 6x – 8? {y|y ≥ –1} {y|y ≤ –1} {y|y ≥ –11} {y|y ≤ –11}

The range of the function f(x)=3x²+6x-8 is:

{y|y ≥ –11}

Step-by-step explanation:

The function f(x)=3x²+6x-8 could also be written as:

f(x)=3x²+6x+3-11

f(x)=3(x²+2x+1)-11

f(x)=3(x+1)²-11

as we know that the value of 3(x+1)²≥0

Then, the value of 3(x+1)²-11≥-11

i.e. f(x)≥-11

Hence, the range of the function f(x)=3x²+6x-8 is:

{y|y ≥ –11}

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!

answer: the answer is 5/22

step-by-step explanation: if you subtract 17 from 22 the answer is 5 therefore the remainder of students are female