Question

h(x) = \

$$\begin{cases} -\\dfrac{1}{2}x + 2 &, -6 \\leq x < 4 \\\\ 2x - 8 &, 4 \\leq x \\leq 8 \\end{cases}$$

what is the graph of h?
choose 1 answer:
a (graph a description)
b (graph b description)

Explanation:

Step1: Analyze the first piece of the function

For \( h(x) = -\frac{1}{2}x + 2 \) with \( -6\leq x < 4 \), we find the value at \( x=-6 \):
\( h(-6)=-\frac{1}{2}(-6)+2 = 3 + 2 = 5 \). So the left - hand part of the piece - wise function should have a point at \( x = - 6,y = 5 \).

Step2: Analyze the second piece of the function

For \( h(x)=2x - 8 \) with \( 4\leq x\leq8 \), we find the value at \( x = 8 \):
\( h(8)=2(8)-8=16 - 8 = 8 \). And at \( x = 4 \), \( h(4)=2(4)-8 = 0 \). Also, for the first piece, when \( x = 4 \) (not included in the first piece's domain, but we can check the limit), \( h(4^-)=-\frac{1}{2}(4)+2=-2 + 2 = 0 \), which matches the value of the second piece at \( x = 4 \).

Now, looking at the two graphs:

• In graph A, the left - most point of the first line segment is at \( x=-6,y = 6 \) (incorrect, since we calculated \( h(-6) = 5 \)).
• In graph B, the left - most point of the first line segment is at \( x=-6,y = 5 \) (correct), and the right - most point of the second line segment at \( x = 8,y = 8 \) (correct), and the point at \( x = 4 \) is \( y = 0 \) for both pieces.

Answer:

B