Graphing quadratic equations worksheet example from a math workbook.

The task involves graphing quadratic equations. Let's analyze and solve the problem step by step for each equation provided.

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Equation 1: \( y = -\frac{1}{3}x^2 + 3 \)

#### Step 1: Identify the type of parabola
- The coefficient of \( x^2 \) is \( -\frac{1}{3} \), which is negative. This means the parabola opens downward.

#### Step 2: Find the vertex
The general form of a quadratic equation is \( y = ax^2 + bx + c \). Here:
- \( a = -\frac{1}{3} \)
- \( b = 0 \)
- \( c = 3 \)

The vertex of a parabola given by \( y = ax^2 + bx + c \) is at:
\[
x = -\frac{b}{2a}
\]
Substitute \( a = -\frac{1}{3} \) and \( b = 0 \):
\[
x = -\frac{0}{2 \left( -\frac{1}{3} \right)} = 0
\]
To find the \( y \)-coordinate of the vertex, substitute \( x = 0 \) into the equation:
\[
y = -\frac{1}{3}(0)^2 + 3 = 3
\]
Thus, the vertex is \( (0, 3) \).

#### Step 3: Find additional points
Choose some \( x \)-values and calculate the corresponding \( y \)-values:
- For \( x = -3 \):
\[
y = -\frac{1}{3}(-3)^2 + 3 = -\frac{1}{3}(9) + 3 = -3 + 3 = 0
\]
Point: \( (-3, 0) \)

- For \( x = 3 \):
\[
y = -\frac{1}{3}(3)^2 + 3 = -\frac{1}{3}(9) + 3 = -3 + 3 = 0
\]
Point: \( (3, 0) \)

- For \( x = -6 \):
\[
y = -\frac{1}{3}(-6)^2 + 3 = -\frac{1}{3}(36) + 3 = -12 + 3 = -9
\]
Point: \( (-6, -9) \)

- For \( x = 6 \):
\[
y = -\frac{1}{3}(6)^2 + 3 = -\frac{1}{3}(36) + 3 = -12 + 3 = -9
\]
Point: \( (6, -9) \)

#### Step 4: Plot the points and sketch the parabola
- Vertex: \( (0, 3) \)
- Points: \( (-3, 0) \), \( (3, 0) \), \( (-6, -9) \), \( (6, -9) \)

The parabola opens downward, with the vertex at \( (0, 3) \) and passing through the points calculated.

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Equation 2: \( y = x^2 - 1\frac{1}{3}x - 7 \)

#### Step 1: Rewrite the equation
Convert the mixed number \( 1\frac{1}{3} \) to an improper fraction:
\[
1\frac{1}{3} = \frac{4}{3}
\]
So the equation becomes:
\[
y = x^2 - \frac{4}{3}x - 7
\]

#### Step 2: Identify the type of parabola
- The coefficient of \( x^2 \) is \( 1 \), which is positive. This means the parabola opens upward.

#### Step 3: Find the vertex
The general form is \( y = ax^2 + bx + c \). Here:
- \( a = 1 \)
- \( b = -\frac{4}{3} \)
- \( c = -7 \)

The vertex is at:
\[
x = -\frac{b}{2a} = -\frac{-\frac{4}{3}}{2(1)} = \frac{\frac{4}{3}}{2} = \frac{4}{3} \cdot \frac{1}{2} = \frac{2}{3}
\]
To find the \( y \)-coordinate of the vertex, substitute \( x = \frac{2}{3} \) into the equation:
\[
y = \left( \frac{2}{3} \right)^2 - \frac{4}{3} \left( \frac{2}{3} \right) - 7
\]
\[
y = \frac{4}{9} - \frac{8}{9} - 7
\]
\[
y = \frac{4}{9} - \frac{8}{9} - \frac{63}{9} = \frac{4 - 8 - 63}{9} = \frac{-67}{9}
\]
Thus, the vertex is \( \left( \frac{2}{3}, -\frac{67}{9} \right) \).

#### Step 4: Find additional points
Choose some \( x \)-values and calculate the corresponding \( y \)-values:
- For \( x = 0 \):
\[
y = (0)^2 - \frac{4}{3}(0) - 7 = -7
\]
Point: \( (0, -7) \)

- For \( x = 3 \):
\[
y = (3)^2 - \frac{4}{3}(3) - 7 = 9 - 4 - 7 = -2
\]
Point: \( (3, -2) \)

- For \( x = -3 \):
\[
y = (-3)^2 - \frac{4}{3}(-3) - 7 = 9 + 4 - 7 = 6
\]
Point: \( (-3, 6) \)

#### Step 5: Plot the points and sketch the parabola
- Vertex: \( \left( \frac{2}{3}, -\frac{67}{9} \right) \)
- Points: \( (0, -7) \), \( (3, -2) \), \( (-3, 6) \)

The parabola opens upward, with the vertex at \( \left( \frac{2}{3}, -\frac{67}{9} \right) \) and passing through the points calculated.

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Final Answer

The graphs of the quadratic equations are sketched based on the vertices and additional points calculated. The final answer is:

\[
\boxed{\text{Graphs are sketched as described above.}}
\]