Algebra 2 Practice - Graph the Quadratic Parabola y = x^2 - 2x - 3 on a Coordinate Plane

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To graph the quadratic parabola \( y = x^2 - 2x - 3 \) on a coordinate plane, follow these steps:

Step 1: Identify the Key Features
The equation is in standard form \( y = ax^2 + bx + c \), where:
\( a = 1 \)
\( b = -2 \)
\( c = -3 \)

Step 2: Find the Vertex
To find the vertex, use the formula for the \( x \)-coordinate of the vertex:
\[
x_{\text{vertex}} = \frac{-b}{2a}
\]
Substitute \( a = 1 \) and \( b = -2 \) into the formula:
\[
x_{\text{vertex}} = \frac{-(-2)}{2(1)} = \frac{2}{2} = 1
\]
Now, substitute \( x = 1 \) into the original equation to find the \( y \)-coordinate of the vertex:
\[
y = (1)^2 - 2(1) - 3 = 1 - 2 - 3 = -4
\]
So, the vertex is \( (1, -4) \).

Step 3: Find the Axis of Symmetry
The axis of symmetry is the vertical line that passes through the \( x \)-coordinate of the vertex. In this case, the axis of symmetry is:
\[
x = 1
\]

Step 4: Plot the Vertex and Axis of Symmetry
Plot the vertex \( (1, -4) \) on the coordinate plane, and draw the axis of symmetry, which is the vertical line \( x = 1 \).

Step 5: Find Additional Points
To get more points for the graph, choose values for \( x \) around the vertex (e.g., \( x = 0 \), \( x = 2 \), and \( x = -1 \)), and substitute them into the equation to find the corresponding \( y \)-values.

For \( x = 0 \):
\[
y = (0)^2 - 2(0) - 3 = -3
\]
So, the point is \( (0, -3) \).

For \( x = 2 \):
\[
y = (2)^2 - 2(2) - 3 = 4 - 4 - 3 = -3
\]
So, the point is \( (2, -3) \).

For \( x = -1 \):
\[
y = (-1)^2 - 2(-1) - 3 = 1 + 2 - 3 = 0
\]
So, the point is \( (-1, 0) \).

Step 6: Plot Additional Points
Plot the points \( (0, -3) \), \( (2, -3) \), and \( (-1, 0) \).

Step 7: Draw the Parabola
Now that you have several points, draw a smooth curve through the points, forming a U-shaped parabola that opens upwards (since \( a = 1 \), which is positive).

Final Graph Features:
Vertex: \( (1, -4) \)
Axis of symmetry: \( x = 1 \)
Parabola opens upwards
The parabola passes through the points \( (-1, 0) \), \( (0, -3) \), \( (2, -3) \), and others along the curve.

The graph of the quadratic parabola should look like a U-shaped curve with the vertex at \( (1, -4) \).

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Nick Perich
Norristown Area High School
Norristown Area School District
Norristown, Pa

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