UpStudy Free Solution:

Let's address each part of the question step-by-step.

(a) Sketch the graph of \(y = | 3x - 2a| \)

The function \(y = | 3x - 2a| \) is an absolute value function, which creates a V-shaped graph. The expression inside the absolute value, \(3x - 2a\), determines the points of intersection with the axes.

Points where the graph cuts or meets the coordinate axes:

1. Y-axis (x = 0):

\[y = | 3( 0) - 2a| = | - 2a | = 2a\]

So, the point on the y-axis is \(( 0, 2a) \).

2. X-axis (y = 0):

\[| 3x - 2a| = 0 \implies 3x - 2a = 0\]

\[3x = 2a \implies x = \frac { 2a} { 3} \]

So, the points on the x-axis are \(\left ( \frac { 2a} { 3} , 0 \right ) \).

Sketch:

The graph of \(y = | 3x - 2a| \) is a V-shaped graph that intersects the y-axis at \(( 0, 2a) \) and the x-axis at \(\left ( \frac { 2a} { 3} , 0 \right ) \).

(b) Solve the inequality \(| 3x - 2a| \leq x + a\)

To solve \(| 3x - 2a| \leq x + a\), we need to consider two cases due to the absolute value.

Case 1: \(3x - 2a \geq 0\)

\[3x - 2a \leq x + a\]

\[3x - x \leq a + 2a\]

\[2x \leq 3a\]

\[x \leq \frac { 3a} { 2} \]

Case 2: \(3x - 2a < 0\)

\[- ( 3x - 2a) \leq x + a\]

\[- 3x + 2a \leq x + a\]

\[2a - a \leq x + 3x\]

\[a \leq 4x\]

\[\frac { a} { 4} \leq x\]

Combining both cases:

\[\frac { a} { 4} \leq x \leq \frac { 3a} { 2} \]

(c) Find the range of possible values of \(g( x) \), where \(g( x) = 5a - \left | \frac { a} { 2} - x \right | \)

To find the range of \(g( x) = 5a - \left | \frac { a} { 2} - x \right | \), we need to analyze the behavior of the absolute value function \(\left | \frac { a} { 2} - x \right | \).

Consider two cases for \(\left | \frac { a} { 2} - x \right | \):

1. When \(x \leq \frac { a} { 2} \):

\[\left | \frac { a} { 2} - x \right | = \frac { a} { 2} - x\]

\[g( x) = 5a - \left ( \frac { a} { 2} - x \right ) \]

\[g( x) = 5a - \frac { a} { 2} + x\]

\[g( x) = \frac { 10a} { 2} - \frac { a} { 2} + x\]

\[g( x) = \frac { 9a} { 2} + x\]

2. When \(x > \frac { a} { 2} \):

\[\left | \frac { a} { 2} - x \right | = x - \frac { a} { 2} \]

\[g( x) = 5a - \left ( x - \frac { a} { 2} \right ) \]

\[g( x) = 5a - x + \frac { a} { 2} \]

\[g( x) = \frac { 10a} { 2} - x + \frac { a} { 2} \]

\[g( x) = \frac { 11a} { 2} - x\]

Range of \(g( x) \):

- For \(x \leq \frac { a} { 2} \):

\[g( x) = \frac { 9a} { 2} + x\]

- For \(x > \frac { a} { 2} \):

\[g( x) = \frac { 11a} { 2} - x\]

The minimum value of \(g( x) \) occurs at \(x = \frac { a} { 2} \):

\[g\left ( \frac { a} { 2} \right ) = 5a - \left | \frac { a} { 2} - \frac { a} { 2} \right | = 5a\]

The maximum value of \(g( x) \) occurs at the endpoints of the interval \(\left [ \frac { a} { 4} , \frac { 3a} { 2} \right ] \):

- At \(x = \frac { a} { 4} \):

\[g\left ( \frac { a} { 4} \right ) = \frac { 9a} { 2} + \frac { a} { 4} = \frac { 18a} { 4} + \frac { a} { 4} = \frac { 19a} { 4} \]

- At \(x = \frac { 3a} { 2} \):

\[g\left ( \frac { 3a} { 2} \right ) = \frac { 11a} { 2} - \frac { 3a} { 2} = \frac { 8a} { 2} = 4a\]

Therefore, the range of \(g( x) \) is:

\[[ 4a, \frac { 19a} { 4} ] \]

Supplemental Knowledge

Absolute value functions and inequalities are fundamental concepts in algebra. They involve expressions that measure the distance of a number from zero on the number line, always yielding a non-negative result.

Key Concepts:

1. Absolute Value Function \(y = | ax + b| \):

- The graph is V-shaped.

- It intersects the y-axis at \(y = | b| \).

- It intersects the x-axis at points where \(ax + b = 0\).

2. Solving Absolute Value Inequalities:

- To solve \(| ax + b| \leq c\), consider two cases:

- \(ax + b \leq c\)

- \(- ( ax + b) \leq c\) or equivalently \(ax + b \geq - c\)

3. Range of Functions Involving Absolute Values:

- Analyze the behavior of the function by considering different intervals based on critical points where the expression inside the absolute value equals zero.

Mastering concepts such as absolute value functions and inequalities is an invaluable way to advance your mathematical ability and confidence. UpStudy can assist with any difficult math problems you encounter if additional help is required!