Question

f(x) = \

$$\begin{cases} \\frac{1}{2}x - 4, & -4 \\leq x \\leq 4 \\\\ 2x - 7, & 4 < x \\leq 6 \\end{cases}$$

what is the graph of f?
choose 1 answer:
a (with graph)
b (with graph)

Explanation:

Step1: Analyze the first piece of the function

The first piece is \( f(x)=\frac{1}{2}x - 4 \) with the domain \( - 4\leq x\leq4 \).

• When \( x=-4 \), \( f(-4)=\frac{1}{2}\times(-4)-4=-2 - 4=-6 \), so the point is \((-4,-6)\) (closed circle since \( x = - 4 \) is included).
• When \( x = 4 \), \( f(4)=\frac{1}{2}\times4-4=2 - 4=-2 \), so the point is \((4,-2)\) (closed circle since \( x = 4 \) is included in this piece).

The slope of this line is \( \frac{1}{2} \), so it's a line with a positive slope, starting at \((-4,-6)\) and ending at \((4,-2)\).

Step2: Analyze the second piece of the function

The second piece is \( f(x)=2x - 7 \) with the domain \( 4\lt x\leq6 \).

• When \( x = 4 \) (not included in this piece), \( f(4)=2\times4-7=8 - 7 = 1 \), so there is an open circle at \((4,1)\).
• When \( x=6 \), \( f(6)=2\times6-7=12 - 7 = 5 \), so the point is \((6,5)\) (closed circle since \( x = 6 \) is included).

The slope of this line is \( 2 \), which is steeper than the slope of the first line.

Now let's analyze the options:

• Option A: The first line has a closed circle at \( x=-4 \) (y=-6), closed circle at \( x = 4 \) (y=-2), and the second line has an open circle at \( x = 4 \) (y = 1) and closed circle at \( x=6 \) (y = 5). The slopes also match (first slope \( \frac{1}{2} \), second slope \( 2 \)).
• Option B: The points and slopes do not match the calculated values for the function. For example, the first line in option B does not seem to pass through \((-4,-6)\) and \((4,-2)\) correctly.

Answer:

A