Question

lesson 15: quadratic functions and models
use the above scatter plot to decide which model fits the data well.
quadratic neither linear
let ( c(t) ) be the expenditures in billions of dollars on dental services for ( t ) years since 1980.
a quadratic model for the data is ( c(t) = square ). use three decimals in your answer.
estimate the expenditures in 2012. ( $ square ). use one decimal in your answer.

Explanation:

Step1: Identify data points

From the scatter plot, the points are $(0,15)$, $(3,20)$, $(6,25)$, $(9,31)$, $(12,39)$, $(15,45)$

Step2: Set up quadratic system

Quadratic form: $C(t)=at^2+bt+c$. Substitute $t=0$:
$C(0)=a(0)^2+b(0)+c=15 \implies c=15$
Substitute $(3,20)$:
$20 = 9a + 3b + 15 \implies 9a+3b=5$ --- (1)
Substitute $(6,25)$:
$25 = 36a + 6b + 15 \implies 36a+6b=10$ --- (2)

Step3: Solve the linear system

Multiply (1) by 2: $18a+6b=10$ --- (3)
Subtract (3) from (2):
$36a+6b-(18a+6b)=10-10 \implies 18a=0 \implies a=0$
Substitute $a=0$ into (1):
$9(0)+3b=5 \implies b=\frac{5}{3}\approx1.667$
Correction: Adjust for curve trend, use least squares
For points $(0,15),(3,20),(6,25),(9,31),(12,39),(15,45)$
Least squares for $C(t)=at^2+bt+c$:
Sum of $t=0+3+6+9+12+15=45$
Sum of $t^2=0+9+36+81+144+225=495$
Sum of $t^3=0+27+216+729+1728+3375=6075$
Sum of $t^4=0+81+1296+6561+20736+50625=79299$
Sum of $C(t)=15+20+25+31+39+45=175$
Sum of $tC(t)=0+60+150+279+468+675=1632$
Sum of $t^2C(t)=0+180+900+2511+5616+10125=19332$

Set up equations:

1. $6c + 45b + 495a = 175$
2. $45c + 495b + 6075a = 1632$
3. $495c + 6075b + 79299a = 19332$

From equation 1: $6c=175-45b-495a \implies c=\frac{175-45b-495a}{6}$
Substitute $c$ into equation 2:
$45\times\frac{175-45b-495a}{6} + 495b + 6075a = 1632$
Multiply through by 6:
$45(175-45b-495a)+2970b+36450a=9792$
$7875-2025b-22275a+2970b+36450a=9792$
$945b+14175a=1917$ --- (4)

Substitute $c$ into equation 3:
$495\times\frac{175-45b-495a}{6} + 6075b + 79299a = 19332$
Multiply through by 6:
$495(175-45b-495a)+36450b+475794a=115992$
$86625-22275b-245025a+36450b+475794a=115992$
$14175b+230769a=29367$ --- (5)

Multiply (4) by 15: $14175b+212625a=28755$ --- (6)
Subtract (6) from (5):
$14175b+230769a-(14175b+212625a)=29367-28755$
$18144a=612 \implies a=\frac{612}{18144}\approx0.034$
Substitute $a=0.034$ into (4):
$945b+14175(0.034)=1917$
$945b+481.95=1917 \implies 945b=1435.05 \implies b\approx1.519$
Substitute $a,b$ into $c$:
$c=\frac{175-45(1.519)-495(0.034)}{6}=\frac{175-68.355-16.83}{6}=\frac{89.815}{6}\approx14.969$

Step4: Write quadratic model

$C(t)=0.034t^2+1.519t+14.969$

Step5: Calculate 2012 expenditure

2012 is $t=2012-1980=32$
$C(32)=0.034(32)^2+1.519(32)+14.969$
$=0.034(1024)+48.608+14.969$
$=34.816+48.608+14.969=98.393\approx98.4$

Answer:

$C(t)=0.034t^2+1.519t+14.969$
Estimated expenditures in 2012: $\$98.4$ billion