Question

graph the function.

\
(x) = \

$$\begin{cases} -x^2 + 1 & \\text{for } x \\le 1 \\\\ 5x - 5 & \\text{for } x > 1 \\end{cases}$$

\\

part 1 of 5
the first rule defines a parabola with vertex \\((0, 1)\\).

part 2 of 5
since \\(x \le 1\\), the rightmost point \\((1, 0)\\) is graphed as a solid dot.

part 3 of 5
the second rule defines a line with slope \\(5\\) and y-intercept \\((0, -5)\\).

part 4 of 5
since \\(x > 1\\), the select point \\((\quad, \quad)\\) is graphed as an open dot.

Explanation:

Analyze the second rule of the piecewise function

The piecewise function is defined as:

$$ r(x) =
LATEXBLOCK0
$$

We are asked to complete Part 4 of 5, which focuses on the boundary point of the second rule, \(5x - 5\) for \(x > 1\).

Determine the boundary point properties

Using the Piecewise Functions and Graphing Piecewise Functions knowledge points:

1. The domain for the second rule is \(x > 1\).
2. The boundary value of \(x\) is \(1\). Since the inequality is strict (\(x > 1\)), this boundary point is the leftmost point of this piece and must be graphed as an open dot.
3. To find the coordinates of this boundary point, we evaluate the rule \(5x - 5\) at \(x = 1\):

$$ y = 5(1) - 5 = 0 $$

Thus, the leftmost point is \((1, 0)\).

Answer:

Since \(x > 1\), the <blank>leftmost</blank> point <blank>\((1, 0)\)</blank> is graphed as an open dot.