This is a homeork problem from my class
"the life expectancy for females born during 1980-1985 was aproximately 77.6 years. this grew to 78 years during 1985-1990 and to 78.6 years during 1990-1995. construct a model for this data by finding a quadratic function whose graph passes through the points (0,77.6), (5,78), and (10,78.6). use this model to estimate the life expectancy for females born between 1995 and 2000 and for those born between 2000 and 2005."
must show all work and use elimination method

Solutions 

To solve this problem we have to use the quadratic function. An equation where the highest exponent is (usually "x") is a square (2).

Quadratic function = f(x)=ax²+bx+c

Three points have been given to us

⇒ (0 , 77.6)      
⇒ (5 ,  78)
⇒ (10 , 78.6)

Calculations 

(0 , 77.6)

x=0
f(x)=77.6

a(0)²+b(0)+c=77.6
c=77.6

(5 , 78)
a(5)²+b(5)+77.6=78
25a+5b=78-77.6
25a+5b=0.4 (1)

(10 , 78.6)
a(10)²+b(10)+77.6=78.6
100a+10b=78.6-77.6
100a+10b=1 (2)

As you see equations number (1) and (2) have a system

100a+10b=1
    25a + 5b=0.4

We solve this system we have to use elimination.  In the elimination method you can add or subtract the equations to get an equation into one variable.

   100a+10b=1
-4(25a+5b=0.4)
≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡
           -10b=-0.6     
 

⇒    b=-0.6/-10=0.06

   100a+10b=1
-2(25a+5b=0.4)
≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡
     50a =  0.2         
 
⇒ a=0.2/50=0.004

Now we have a, b and c:

α = 0.004
β = 0.06
c = 77.6

Therefore, the quadratic function is:

f(x)=0.004x²+0.06x+77.6

x=year of the beginning of the interval - 1980

The life expectancy for females born between 1995 and 2000 is when

x=1995-1980=15

Therefore we do 

f(15)=0.004(15)²+0.06(15)+77.6
f(15)=0.004(225)+0.06(15)+77.6
f(15)=79.4

Females between 2000 and 2005 

x=2000-1980=20

Now we do 

f(20)=0.004(20)²+0.06(20)+77.6

f(20)=1.6+1.2+77.6

f(20)=80.4

The life expectancy for females born between 1995 and 2000 = 79.4 years.

The life expectancy for females born between 2000 and 2005 = 80.4 years.

This is a quadratic function:

f(x)=ax²+bx+c

We have three points:
(0 , 77.6)     
(5 ,  78)
(10 , 78.6)

Then, we make the function wiht these points:
(0 , 77.6)

x=0
f(x)=77.6

a(0)²+b(0)+c=77.6
c=77.6

(5 , 78)
a(5)²+b(5)+77.6=78
25a+5b=78-77.6
25a+5b=0.4         (1)

(10 , 78.6)
a(10)²+b(10)+77.6=78.6
100a+10b=78.6-77.6
100a+10b=1        (2)

With the equations (1) and (2) we have a system of equations:
100a+10b=1
  25a + 5b=0.4

We solve this system of equations by method of elimination.

   100a+10b=1
-4(25a+5b=0.4)

           -10b=-0.6      ⇒    b=-0.6/-10=0.06

   100a+10b=1
-2(25a+5b=0.4)

     50a =  0.2          ⇒ a=0.2/50=0.004

We have a, b and c:
a=0.004
b=0.06
c=77.6

Therefore, the quadratic funtion is:
f(x)=0.004x²+0.06x+77.6
x=year of the beginning of the interval - 1980

The life expectancy for females born between 1995 and 2000 is when x=1995-1980=15
Therefore:
f(15)=0.004(15)²+0.06(15)+77.6
f(15)=0.004(225)+0.06(15)+77.6
f(15)=79.4

The life expectancy for females born between 2000 and 2005 is when:
x=2000-1980=20
therefore:
f(20)=0.004(20)²+0.06(20)+77.6
f(20)=1.6+1.2+77.6
f(20)=80.4

The funtion is:
f(x)=0.004x²+0.06x+77.6
x=year of the beginning of the interval - 1980

The life expectancy for females born between 1995 and 2000 is 79.4 years.
The life expectancy for females born between 2000 and 2005 is 80.4 years.

The answer is
k<4
Hope that helps!