Sure! Let's work through each part of the problem step-by-step.
### Part a:
We are given the function [tex]\( f(x) = x^2 + \frac{9}{2} x \)[/tex].
#### Step 1: Evaluate [tex]\( f(2) \)[/tex]
Plug [tex]\( x = 2 \)[/tex] into the function:
[tex]\[
f(2) = 2^2 + \frac{9}{2} \cdot 2
\][/tex]
Simplify the terms inside the function:
[tex]\[
f(2) = 4 + 9 = 13
\][/tex]
So, [tex]\( f(2) = 13 \)[/tex].
#### Step 2: Solve for [tex]\( x \)[/tex] when [tex]\( f(x) = 17 \frac{1}{2} \)[/tex]
First, convert the mixed number to an improper fraction:
[tex]\[
17 \frac{1}{2} = 17.5
\][/tex]
We need to solve the equation [tex]\( f(x) = 17.5 \)[/tex]:
[tex]\[
x^2 + \frac{9}{2} x = 17.5
\][/tex]
Rearrange the equation to set it to zero:
[tex]\[
x^2 + \frac{9}{2} x - 17.5 = 0
\][/tex]
Solving this quadratic equation, we find the solutions:
[tex]\[
x = -7 \quad \text{and} \quad x = 2.5
\][/tex]
So, the values of [tex]\( x \)[/tex] that satisfy [tex]\( f(x) = 17.5 \)[/tex] are [tex]\( x = -7 \)[/tex] and [tex]\( x = 2.5 \)[/tex].
### Part b:
We are given the linear function [tex]\( f(x) = ax + b \)[/tex] and two points on the line: [tex]\( f(2) = 1 \)[/tex] and [tex]\( f(4) = 3 \)[/tex].
#### Step 1: Set up the system of equations
From the given points:
1. When [tex]\( x = 2 \)[/tex], [tex]\( f(2) = 1 \)[/tex]:
[tex]\[
2a + b = 1
\][/tex]
2. When [tex]\( x = 4 \)[/tex], [tex]\( f(4) = 3 \)[/tex]:
[tex]\[
4a + b = 3
\][/tex]
#### Step 2: Solve the system of equations
We can subtract the first equation from the second to eliminate [tex]\( b \)[/tex]:
[tex]\[
(4a + b) - (2a + b) = 3 - 1
\][/tex]
[tex]\[
2a = 2
\][/tex]
[tex]\[
a = 1
\][/tex]
Now, substitute [tex]\( a = 1 \)[/tex] back into the first equation:
[tex]\[
2(1) + b = 1
\][/tex]
[tex]\[
2 + b = 1
\][/tex]
[tex]\[
b = -1
\][/tex]
So, the linear function is:
[tex]\[
f(x) = 1x - 1
\][/tex]
[tex]\[
f(x) = x - 1
\][/tex]
Therefore, the function [tex]\( f(x) \)[/tex] is [tex]\( f(x) = x - 1 \)[/tex].
### Summary:
For part a, [tex]\( f(2) = 13 \)[/tex] and the values of [tex]\( x \)[/tex] such that [tex]\( f(x) = 17.5 \)[/tex] are [tex]\( x = -7 \)[/tex] and [tex]\( x = 2.5 \)[/tex].
For part b, the linear function [tex]\( f(x) \)[/tex] is [tex]\( f(x) = x - 1 \)[/tex].
