Worksheet for graphing quadratic inequalities, featuring four problems with corresponding graphs and shaded regions.

To solve the problem of graphing quadratic inequalities, we need to follow a systematic approach for each inequality. Let's go through each one step by step.

1) \( y > 3x^2 - 12x + 11 \)

#### Step 1: Graph the quadratic equation \( y = 3x^2 - 12x + 11 \)
- Vertex Form: First, find the vertex of the parabola.
- The vertex \( x \)-coordinate is given by \( x = -\frac{b}{2a} \).
\[
x = -\frac{-12}{2 \cdot 3} = \frac{12}{6} = 2
\]
- Substitute \( x = 2 \) into the equation to find the \( y \)-coordinate:
\[
y = 3(2)^2 - 12(2) + 11 = 3(4) - 24 + 11 = 12 - 24 + 11 = -1
\]
So, the vertex is \( (2, -1) \).

- Plot Points: Choose a few \( x \)-values to find corresponding \( y \)-values:
- \( x = 0 \): \( y = 3(0)^2 - 12(0) + 11 = 11 \)
- \( x = 1 \): \( y = 3(1)^2 - 12(1) + 11 = 3 - 12 + 11 = 2 \)
- \( x = 3 \): \( y = 3(3)^2 - 12(3) + 11 = 27 - 36 + 11 = 2 \)
- \( x = 4 \): \( y = 3(4)^2 - 12(4) + 11 = 48 - 48 + 11 = 11 \)

- Graph the Parabola: Plot the points and draw a parabola opening upwards (since the coefficient of \( x^2 \) is positive).

#### Step 2: Determine the Region
- The inequality is \( y > 3x^2 - 12x + 11 \), which means we shade the region above the parabola.

#### Final Graph:
- Draw the parabola as a dashed line (since the inequality is strict, \( > \)).
- Shade the region above the parabola.

2) \( y < x^2 - 6x + 11 \)

#### Step 1: Graph the quadratic equation \( y = x^2 - 6x + 11 \)
- Vertex Form: Find the vertex.
- The vertex \( x \)-coordinate is:
\[
x = -\frac{-6}{2 \cdot 1} = \frac{6}{2} = 3
\]
- Substitute \( x = 3 \) into the equation:
\[
y = (3)^2 - 6(3) + 11 = 9 - 18 + 11 = 2
\]
So, the vertex is \( (3, 2) \).

- Plot Points: Choose a few \( x \)-values:
- \( x = 0 \): \( y = (0)^2 - 6(0) + 11 = 11 \)
- \( x = 1 \): \( y = (1)^2 - 6(1) + 11 = 1 - 6 + 11 = 6 \)
- \( x = 2 \): \( y = (2)^2 - 6(2) + 11 = 4 - 12 + 11 = 3 \)
- \( x = 4 \): \( y = (4)^2 - 6(4) + 11 = 16 - 24 + 11 = 3 \)
- \( x = 5 \): \( y = (5)^2 - 6(5) + 11 = 25 - 30 + 11 = 6 \)

- Graph the Parabola: Plot the points and draw a parabola opening upwards.

#### Step 2: Determine the Region
- The inequality is \( y < x^2 - 6x + 11 \), so we shade the region below the parabola.

#### Final Graph:
- Draw the parabola as a dashed line.
- Shade the region below the parabola.

3) \( y > -4x^2 - 24x - 38 \)

#### Step 1: Graph the quadratic equation \( y = -4x^2 - 24x - 38 \)
- Vertex Form: Find the vertex.
- The vertex \( x \)-coordinate is:
\[
x = -\frac{-24}{2 \cdot -4} = \frac{24}{-8} = -3
\]
- Substitute \( x = -3 \) into the equation:
\[
y = -4(-3)^2 - 24(-3) - 38 = -4(9) + 72 - 38 = -36 + 72 - 38 = -2
\]
So, the vertex is \( (-3, -2) \).

- Plot Points: Choose a few \( x \)-values:
- \( x = -4 \): \( y = -4(-4)^2 - 24(-4) - 38 = -4(16) + 96 - 38 = -64 + 96 - 38 = -4 \)
- \( x = -2 \): \( y = -4(-2)^2 - 24(-2) - 38 = -4(4) + 48 - 38 = -16 + 48 - 38 = -6 \)
- \( x = -1 \): \( y = -4(-1)^2 - 24(-1) - 38 = -4(1) + 24 - 38 = -4 + 24 - 38 = -18 \)
- \( x = 0 \): \( y = -4(0)^2 - 24(0) - 38 = -38 \)

- Graph the Parabola: Plot the points and draw a parabola opening downwards (since the coefficient of \( x^2 \) is negative).

#### Step 2: Determine the Region
- The inequality is \( y > -4x^2 - 24x - 38 \), so we shade the region above the parabola.

#### Final Graph:
- Draw the parabola as a dashed line.
- Shade the region above the parabola.

4) \( y < -4x^2 - 1 \)

#### Step 1: Graph the quadratic equation \( y = -4x^2 - 1 \)
- Vertex Form: The vertex is at \( (0, -1) \) because there is no \( x \)-term.
- Plot Points: Choose a few \( x \)-values:
- \( x = -1 \): \( y = -4(-1)^2 - 1 = -4(1) - 1 = -4 - 1 = -5 \)
- \( x = 1 \): \( y = -4(1)^2 - 1 = -4(1) - 1 = -4 - 1 = -5 \)
- \( x = -2 \): \( y = -4(-2)^2 - 1 = -4(4) - 1 = -16 - 1 = -17 \)
- \( x = 2 \): \( y = -4(2)^2 - 1 = -4(4) - 1 = -16 - 1 = -17 \)

- Graph the Parabola: Plot the points and draw a parabola opening downwards.

#### Step 2: Determine the Region
- The inequality is \( y < -4x^2 - 1 \), so we shade the region below the parabola.

#### Final Graph:
- Draw the parabola as a dashed line.
- Shade the region below the parabola.

Final Answer

The graphs for each inequality are as follows:

1. \( y > 3x^2 - 12x + 11 \): Shade above the parabola.
2. \( y < x^2 - 6x + 11 \): Shade below the parabola.
3. \( y > -4x^2 - 24x - 38 \): Shade above the parabola.
4. \( y < -4x^2 - 1 \): Shade below the parabola.

\[
\boxed{
\begin{array}{c}
\text{1) Shade above the parabola} \\
\text{2) Shade below the parabola} \\
\text{3) Shade above the parabola} \\
\text{4) Shade below the parabola}
\end{array}
}
\]