Plotting Quadratic Graphs - GCSE Maths - Steps & Examples - Free Printable

Problem Description:

The task involves plotting a quadratic graph for the function \( y = x^2 + 2x + 5 \). The provided example includes a table of \( x \) and \( y \) values, and a graph is plotted based on these values. The goal is to understand how the quadratic function is evaluated and how the graph is constructed.

Solution Explanation:

#### Step 1: Understanding the Quadratic Function
The given quadratic function is:
\[
y = x^2 + 2x + 5
\]
This is a standard quadratic equation in the form \( y = ax^2 + bx + c \), where:
- \( a = 1 \)
- \( b = 2 \)
- \( c = 5 \)

#### Step 2: Creating a Table of Values
To plot the graph, we need to calculate the corresponding \( y \)-values for a set of \( x \)-values. The table provided in the image uses the following \( x \)-values: \( -3, -2, -1, 0, 1, 2 \).

We substitute each \( x \)-value into the quadratic equation \( y = x^2 + 2x + 5 \) to find the corresponding \( y \)-values.

##### Calculation for Each \( x \)-value:
1. For \( x = -3 \):
\[
y = (-3)^2 + 2(-3) + 5 = 9 - 6 + 5 = 8
\]

2. For \( x = -2 \):
\[
y = (-2)^2 + 2(-2) + 5 = 4 - 4 + 5 = 5
\]

3. For \( x = -1 \):
\[
y = (-1)^2 + 2(-1) + 5 = 1 - 2 + 5 = 4
\]

4. For \( x = 0 \):
\[
y = (0)^2 + 2(0) + 5 = 0 + 0 + 5 = 5
\]

5. For \( x = 1 \):
\[
y = (1)^2 + 2(1) + 5 = 1 + 2 + 5 = 8
\]

6. For \( x = 2 \):
\[
y = (2)^2 + 2(2) + 5 = 4 + 4 + 5 = 13
\]

#### Step 3: Constructing the Table
Using the calculated \( y \)-values, we construct the following table:

| \( x \) | \( -3 \) | \( -2 \) | \( -1 \) | \( 0 \) | \( 1 \) | \( 2 \) |
|---------|----------|----------|----------|---------|---------|---------|
| \( y \) | \( 8 \) | \( 5 \) | \( 4 \) | \( 5 \) | \( 8 \) | \( 13 \) |

#### Step 4: Plotting the Points
Next, we plot the points \((x, y)\) on a Cartesian coordinate system:
- \((-3, 8)\)
- \((-2, 5)\)
- \((-1, 4)\)
- \((0, 5)\)
- \((1, 8)\)
- \((2, 13)\)

#### Step 5: Drawing the Quadratic Curve
After plotting the points, we connect them smoothly to form a parabolic curve. The resulting graph is a upward-opening parabola because the coefficient of \( x^2 \) (which is \( a = 1 \)) is positive.

#### Step 6: Identifying Key Features of the Parabola
- Vertex: The vertex is the lowest point of the parabola. From the table, we see that the minimum \( y \)-value occurs at \( x = -1 \) with \( y = 4 \). Thus, the vertex is \((-1, 4)\).
- Axis of Symmetry: The axis of symmetry is a vertical line passing through the vertex. For a quadratic function \( y = ax^2 + bx + c \), the axis of symmetry is given by \( x = -\frac{b}{2a} \). Here:
\[
x = -\frac{2}{2(1)} = -1
\]
This confirms that the axis of symmetry is \( x = -1 \).

#### Final Graph:
The graph shows a parabola opening upwards, with the vertex at \((-1, 4)\) and the axis of symmetry at \( x = -1 \).

Final Answer:

\[
\boxed{\text{The quadratic graph of } y = x^2 + 2x + 5 \text{ is plotted using the points from the table, forming an upward-opening parabola with vertex at } (-1, 4).}
\]