Question

This is a homeork problem from my class

Answer 1

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

Answer:
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.

Answer 2

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.