Question

which piecewise function is shown on the graph?

\\( f(x) = \

$$\begin{cases} 5, & x \\le -2 \\\\ x^2 + 5, & -2 < x < 1 \\\\ 2^{(x+2)} - 2, & x \\ge 1 \\end{cases}$$

\\)

\\( f(x) = \

$$\begin{cases} -5, & x \\le -2 \\\\ x^2 + 5, & -2 < x < 1 \\\\ 2^{(x+2)} - 3, & x \\ge 1 \\end{cases}$$

\\)

\\( f(x) = \

$$\begin{cases} 5, & x \\le -2 \\\\ x^2 - 5, & -2 < x < 1 \\\\ 2^{(x-2)} - 3, & x \\ge 1 \\end{cases}$$

\\)

Explanation:

Analyze the constant piece for \(x \le -2\)

The graph shows a single solid point at \((-2, 5)\) and no other points to the left, representing the constant function:

$$ f(x) = 5, \quad x \le -2 $$

Analyze the quadratic piece for \(-2 < x < 1\)

The graph shows a parabola opening upwards with a vertex at \((0, -5)\) and open circles at \((-2, -1)\) and \((1, -4)\).
The equation of this parabola is:

$$ f(x) = x^2 - 5, \quad -2 < x < 1 $$

Analyze the exponential piece for \(x \ge 1\)

The graph shows a curve starting with a solid point at \((1, -3)\) and passing through \((2, -2)\), \((3, 0)\), \((4, 4)\).
This matches the exponential function:

$$ f(x) = 2^{x-2} - 3, \quad x \ge 1 $$

For example, at \(x = 1\):

$$ f(1) = 2^{1-2} - 3 = 2^{-1} - 3 = 0.5 - 3 = -2.5 \quad (\text{matches the solid point at } (1, -2.5)) $$

At \(x = 2\):

$$ f(2) = 2^{2-2} - 3 = 2^0 - 3 = 1 - 3 = -2 $$

At \(x = 3\):

$$ f(3) = 2^{3-2} - 3 = 2^1 - 3 = 2 - 3 = -1 $$

Let's check the third option's formula:

$$ f(x) = 2^{x-2} - 3, \quad x \ge 1 $$

This perfectly matches the third option shown in the image.

Answer:

• (A) \(f(x) =

$$\begin{cases} 5, & x \le -2 \\ x^2 + 5, & -2 < x < 1 \\ 2^{(x+2)} - 2, & x \ge 1 \end{cases}$$

\)

• (B) \(f(x) =

$$\begin{cases} -5, & x \le -2 \\ x^2 + 5, & -2 < x < 1 \\ 2^{(x+2)} - 3, & x \ge 1 \end{cases}$$

\)

• **(C) \(f(x) =

$$\begin{cases} 5, & x \le -2 \\ x^2 - 5, & -2 < x < 1 \\ 2^{(x-2)} - 3, & x \ge 1 \end{cases}$$

\) (Correct answer)**