Worksheet for practicing graphing quadratic functions by completing tables and plotting points.

Problem: Graphing Quadratic Functions

The task involves completing tables for six quadratic functions and sketching their graphs. Below, I will solve each part step by step.

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#### 1. \( f(x) = x^2 + 4x - 2 \)

Table:
\[
\begin{array}{c|c}
x & f(x) \\
\hline
0 & f(0) = 0^2 + 4(0) - 2 = -2 \\
1 & f(1) = 1^2 + 4(1) - 2 = 1 + 4 - 2 = 3 \\
2 & f(2) = 2^2 + 4(2) - 2 = 4 + 8 - 2 = 10 \\
3 & f(3) = 3^2 + 4(3) - 2 = 9 + 12 - 2 = 19 \\
4 & f(4) = 4^2 + 4(4) - 2 = 16 + 16 - 2 = 30 \\
\end{array}
\]

Completed Table:
\[
\begin{array}{c|c}
x & f(x) \\
\hline
0 & -2 \\
1 & 3 \\
2 & 10 \\
3 & 19 \\
4 & 30 \\
\end{array}
\]

Graph:
- The function is a parabola opening upwards.
- Vertex can be found using \( x = -\frac{b}{2a} = -\frac{4}{2(1)} = -2 \).
- Plot the points from the table and sketch the parabola.

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#### 2. \( f(x) = x^2 + 8x + 13 \)

Table:
\[
\begin{array}{c|c}
x & f(x) \\
\hline
-6 & f(-6) = (-6)^2 + 8(-6) + 13 = 36 - 48 + 13 = 1 \\
-5 & f(-5) = (-5)^2 + 8(-5) + 13 = 25 - 40 + 13 = -2 \\
-4 & f(-4) = (-4)^2 + 8(-4) + 13 = 16 - 32 + 13 = -3 \\
-3 & f(-3) = (-3)^2 + 8(-3) + 13 = 9 - 24 + 13 = -2 \\
-2 & f(-2) = (-2)^2 + 8(-2) + 13 = 4 - 16 + 13 = 1 \\
\end{array}
\]

Completed Table:
\[
\begin{array}{c|c}
x & f(x) \\
\hline
-6 & 1 \\
-5 & -2 \\
-4 & -3 \\
-3 & -2 \\
-2 & 1 \\
\end{array}
\]

Graph:
- The function is a parabola opening upwards.
- Vertex can be found using \( x = -\frac{b}{2a} = -\frac{8}{2(1)} = -4 \).
- Plot the points from the table and sketch the parabola.

---

#### 3. \( f(x) = x^2 - 2x - 2 \)

Table:
\[
\begin{array}{c|c}
x & f(x) \\
\hline
-1 & f(-1) = (-1)^2 - 2(-1) - 2 = 1 + 2 - 2 = 1 \\
0 & f(0) = 0^2 - 2(0) - 2 = -2 \\
1 & f(1) = 1^2 - 2(1) - 2 = 1 - 2 - 2 = -3 \\
2 & f(2) = 2^2 - 2(2) - 2 = 4 - 4 - 2 = -2 \\
3 & f(3) = 3^2 - 2(3) - 2 = 9 - 6 - 2 = 1 \\
\end{array}
\]

Completed Table:
\[
\begin{array}{c|c}
x & f(x) \\
\hline
-1 & 1 \\
0 & -2 \\
1 & -3 \\
2 & -2 \\
3 & 1 \\
\end{array}
\]

Graph:
- The function is a parabola opening upwards.
- Vertex can be found using \( x = -\frac{b}{2a} = -\frac{-2}{2(1)} = 1 \).
- Plot the points from the table and sketch the parabola.

---

#### 4. \( f(x) = -x(x + 2) \)

First, expand the function:
\[
f(x) = -x(x + 2) = -x^2 - 2x
\]

Table:
\[
\begin{array}{c|c}
x & f(x) \\
\hline
-3 & f(-3) = -(-3)^2 - 2(-3) = -9 + 6 = -3 \\
-2 & f(-2) = -(-2)^2 - 2(-2) = -4 + 4 = 0 \\
-1 & f(-1) = -(-1)^2 - 2(-1) = -1 + 2 = 1 \\
0 & f(0) = -(0)^2 - 2(0) = 0 \\
1 & f(1) = -(1)^2 - 2(1) = -1 - 2 = -3 \\
\end{array}
\]

Completed Table:
\[
\begin{array}{c|c}
x & f(x) \\
\hline
-3 & -3 \\
-2 & 0 \\
-1 & 1 \\
0 & 0 \\
1 & -3 \\
\end{array}
\]

Graph:
- The function is a parabola opening downwards.
- Vertex can be found using \( x = -\frac{b}{2a} = -\frac{-2}{2(-1)} = -1 \).
- Plot the points from the table and sketch the parabola.

---

#### 5. \( f(x) = -x - 4x^2 + 5 \)

Rearrange the function:
\[
f(x) = -4x^2 - x + 5
\]

Table:
\[
\begin{array}{c|c}
x & f(x) \\
\hline
2 & f(2) = -4(2)^2 - (2) + 5 = -4(4) - 2 + 5 = -16 - 2 + 5 = -13 \\
3 & f(3) = -4(3)^2 - (3) + 5 = -4(9) - 3 + 5 = -36 - 3 + 5 = -34 \\
4 & f(4) = -4(4)^2 - (4) + 5 = -4(16) - 4 + 5 = -64 - 4 + 5 = -63 \\
5 & f(5) = -4(5)^2 - (5) + 5 = -4(25) - 5 + 5 = -100 - 5 + 5 = -100 \\
6 & f(6) = -4(6)^2 - (6) + 5 = -4(36) - 6 + 5 = -144 - 6 + 5 = -145 \\
\end{array}
\]

Completed Table:
\[
\begin{array}{c|c}
x & f(x) \\
\hline
2 & -13 \\
3 & -34 \\
4 & -63 \\
5 & -100 \\
6 & -145 \\
\end{array}
\]

Graph:
- The function is a parabola opening downwards.
- Vertex can be found using \( x = -\frac{b}{2a} = -\frac{-1}{2(-4)} = -\frac{1}{8} \).
- Plot the points from the table and sketch the parabola.

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#### 6. \( f(x) = x^2 \)

Table:
\[
\begin{array}{c|c}
x & f(x) \\
\hline
-2 & f(-2) = (-2)^2 = 4 \\
-1 & f(-1) = (-1)^2 = 1 \\
0 & f(0) = (0)^2 = 0 \\
1 & f(1) = (1)^2 = 1 \\
2 & f(2) = (2)^2 = 4 \\
\end{array}
\]

Completed Table:
\[
\begin{array}{c|c}
x & f(x) \\
\hline
-2 & 4 \\
-1 & 1 \\
0 & 0 \\
1 & 1 \\
2 & 4 \\
\end{array}
\]

Graph:
- The function is a standard parabola opening upwards.
- Vertex is at \( (0, 0) \).
- Plot the points from the table and sketch the parabola.

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Final Answer:

\[
\boxed{\text{See detailed solutions and graphs above.}}
\]