Math 2 Piecewise Functions Worksheet 2 - Free Printable

Let's solve each problem step by step.

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Problem 1:

$$
f(x) =
\begin{cases}
x + 5 & \text{if } x < -2 \\
x^2 + 2x + 3 & \text{if } x \geq -2
\end{cases}
$$

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#### Step 1: Determine if it's a function

A piecewise-defined function is still a function as long as each input $ x $ has exactly one output. Here, the two pieces are defined on non-overlapping domains:
- $ x < -2 $ → first rule
- $ x \geq -2 $ → second rule

At $ x = -2 $, only the second rule applies (since $ x \geq -2 $). So no overlap or ambiguity.

✔ Yes, this is a function.

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#### Step 2: Evaluate at specific values

##### a) $ f(3) $

Since $ 3 \geq -2 $, use the second rule:
$$
f(3) = 3^2 + 2(3) + 3 = 9 + 6 + 3 = 18
$$

✔ $ f(3) = 18 $

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##### b) $ f(-4) $

Since $ -4 < -2 $, use the first rule:
$$
f(-4) = -4 + 5 = 1
$$

✔ $ f(-4) = 1 $

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##### c) $ f(-2) $

Since $ -2 \geq -2 $, use the second rule:
$$
f(-2) = (-2)^2 + 2(-2) + 3 = 4 - 4 + 3 = 3
$$

✔ $ f(-2) = 3 $

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#### Step 3: Graphing

We’ll sketch two parts:

1. For $ x < -2 $: $ y = x + 5 $
- This is a straight line with slope 1 and y-intercept 5.
- But only for $ x < -2 $. At $ x = -2 $, it’s not included, so we draw an open circle at $ x = -2 $.
- When $ x = -2 $, $ y = -2 + 5 = 3 $, but since $ x < -2 $, we don’t include that point.
- So plot points like:
- $ x = -3 $: $ y = 2 $
- $ x = -4 $: $ y = 1 $
- $ x = -5 $: $ y = 0 $
- Draw a ray to the left from $ (-2, 3) $, open circle at $ (-2, 3) $, extending left.

2. For $ x \geq -2 $: $ y = x^2 + 2x + 3 $
- This is a parabola.
- Let’s find vertex: $ x = -\frac{b}{2a} = -\frac{2}{2(1)} = -1 $
- $ f(-1) = (-1)^2 + 2(-1) + 3 = 1 - 2 + 3 = 2 $
- So vertex is $ (-1, 2) $
- Include point at $ x = -2 $: $ f(-2) = 3 $ → closed circle at $ (-2, 3) $
- Plot a few more points:
- $ x = 0 $: $ f(0) = 0 + 0 + 3 = 3 $
- $ x = 1 $: $ f(1) = 1 + 2 + 3 = 6 $
- $ x = -3 $: Not in domain (but just for continuity check)

So the graph:
- Left side: line $ y = x+5 $, open at $ (-2, 3) $
- Right side: parabola starting at $ (-2, 3) $, going through $ (-1, 2) $, $ (0, 3) $, $ (1, 6) $, etc.

> ✔ The two graphs meet at $ (-2, 3) $ — the parabola starts there with a closed dot, the line ends there with an open dot, which is correct.

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Answer Summary for Problem 1:

- Function? Yes
- $ f(3) = 18 $
- $ f(-4) = 1 $
- $ f(-2) = 3 $

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Problem 2:

$$
f(x) =
\begin{cases}
2x + 1 & \text{if } x \geq 1 \\
x^2 + 3 & \text{if } x < 1
\end{cases}
$$

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#### Step 1: Is this a function?

Again, the domains are:
- $ x \geq 1 $ → first rule
- $ x < 1 $ → second rule

They cover all real numbers without overlap. At $ x = 1 $, only the first rule applies.

✔ Yes, this is a function.

---

#### Step 2: Evaluate at specific values

##### a) $ f(-2) $

$ -2 < 1 $ → use second rule:
$$
f(-2) = (-2)^2 + 3 = 4 + 3 = 7
$$

✔ $ f(-2) = 7 $

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##### b) $ f(6) $

$ 6 \geq 1 $ → use first rule:
$$
f(6) = 2(6) + 1 = 12 + 1 = 13
$$

✔ $ f(6) = 13 $

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##### c) $ f(1) $

$ 1 \geq 1 $ → use first rule:
$$
f(1) = 2(1) + 1 = 2 + 1 = 3
$$

✔ $ f(1) = 3 $

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#### Step 3: Graphing

1. For $ x < 1 $: $ y = x^2 + 3 $
- Parabola shifted up by 3 units.
- Vertex at $ (0, 3) $
- Only for $ x < 1 $, so at $ x = 1 $, we have an open circle at $ (1, 1^2 + 3) = (1, 4) $
- Plot points:
- $ x = 0 $: $ y = 3 $
- $ x = -1 $: $ y = 1 + 3 = 4 $
- $ x = -2 $: $ y = 4 + 3 = 7 $
- $ x = 0.5 $: $ y = 0.25 + 3 = 3.25 $
- Draw parabola from left up to $ x = 1 $, open circle at $ (1, 4) $

2. For $ x \geq 1 $: $ y = 2x + 1 $
- Straight line with slope 2, y-intercept 1.
- At $ x = 1 $: $ y = 2(1) + 1 = 3 $ → closed circle at $ (1, 3) $
- Plot:
- $ x = 2 $: $ y = 5 $
- $ x = 3 $: $ y = 7 $
- $ x = 0 $: not in domain

So the graph:
- Left part: parabola $ y = x^2 + 3 $, ending with open circle at (1, 4).
- Right part: line $ y = 2x + 1 $, starting at closed circle at (1, 3).
- There is a jump discontinuity at $ x = 1 $: from $ (1, 4) $ (not included) to $ (1, 3) $ (included).

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Answer Summary for Problem 2:

- Function? Yes
- $ f(-2) = 7 $
- $ f(6) = 13 $
- $ f(1) = 3 $

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✔ Final Answers:

#### Problem 1:
- Function? Yes
- $ f(3) = 18 $
- $ f(-4) = 1 $
- $ f(-2) = 3 $

#### Problem 2:
- Function? Yes
- $ f(-2) = 7 $
- $ f(6) = 13 $
- $ f(1) = 3 $

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You can now sketch both graphs on the grids provided using the descriptions above.

Let me know if you'd like a visual description of how to draw them!