To solve this problem, we need to determine a linear function that represents the cost of tuition as a function of [tex]\( x \)[/tex], the number of years since 1990. Additionally, we need to find the tuition cost in the year 2006 and the year when the tuition will be \[tex]$381 per credit hour.
### Step 1: Determine the Linear Function \( C(x) \)
We are given two points:
- In 1990 (which we will call \( x = 0 \)), the tuition was \$[/tex]96.
- In 2003 (which is [tex]\( x = 2003 - 1990 = 13 \)[/tex] years since 1990), the tuition was \[tex]$291.
First, we find the slope of the linear function. The slope \( m \) is given by the change in tuition divided by the change in years:
\[ m = \frac{291 - 96}{13 - 0} = \frac{195}{13} = 15 \]
Next, we use the slope and one of the points to determine the y-intercept \( b \). Using the point (0, 96):
\[ C(0) = m \cdot 0 + b = 96 \implies b = 96 \]
So the linear function \( C(x) \) is:
\[ C(x) = 15x + 96 \]
### Step 2: Calculate Tuition in 2006
To find the tuition in 2006, we identify the number of years since 1990:
\[ x = 2006 - 1990 = 16 \]
We substitute \( x = 16 \) into the linear function \( C(x) \):
\[ C(16) = 15 \cdot 16 + 96 = 240 + 96 = 336 \]
So, in the year 2006, the tuition will be \$[/tex]336 per credit hour.
### Step 3: Determine the Year When Tuition Will Be \[tex]$381
We set the function \( C(x) \) equal to 381 and solve for \( x \):
\[ 15x + 96 = 381 \]
\[ 15x = 381 - 96 \]
\[ 15x = 285 \]
\[ x = \frac{285}{15} = 19 \]
Adding these 19 years to the base year of 1990:
\[ 1990 + 19 = 2009 \]
Therefore, in the year 2009, the tuition will be \$[/tex]381 per credit hour.
### Final Answer
- The linear function [tex]\( C(x) \)[/tex] that represents the cost of tuition is:
[tex]\[ C(x) = 15x + 96 \][/tex]
- In the year 2006, the tuition will be \[tex]$336 per credit hour.
- In the year 2009, the tuition will be \$[/tex]381 per credit hour.
