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To graph the equation [tex]\( y = x^2 + 2x - 8 \)[/tex] on the given set of axes, we'll follow a few key steps: finding the roots, the vertex, and identifying a few additional points for a clear and accurate graph. Let's begin:
1. Find the Roots (Zeros) of the Equation:
The roots are the values of [tex]\( x \)[/tex] for which [tex]\( y = 0 \)[/tex].
For the equation [tex]\( x^2 + 2x - 8 = 0 \)[/tex]:
The roots (or x-intercepts) are:
[tex]\[ x = 2 \quad \text{and} \quad x = -4 \][/tex]
So, the points (roots) are:
[tex]\[ (2, 0) \quad \text{and} \quad (-4, 0) \][/tex]
2. Find the Vertex of the Parabola:
The vertex of a quadratic equation [tex]\( y = ax^2 + bx + c \)[/tex] is given by:
[tex]\[ x = -\frac{b}{2a} \][/tex]
For [tex]\( y = x^2 + 2x - 8 \)[/tex], [tex]\( a = 1 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -8 \)[/tex]. Thus:
[tex]\[ x = -\frac{2}{2 \cdot 1} = -1 \][/tex]
Plugging [tex]\( x = -1 \)[/tex] back into the equation to get the corresponding [tex]\( y \)[/tex]-value:
[tex]\[ y = (-1)^2 + 2(-1) - 8 = 1 - 2 - 8 = -9 \][/tex]
Therefore, the vertex of the parabola is:
[tex]\[ (-1, -9) \][/tex]
3. Choose and Calculate Additional Points:
To have a well-formed graph, we select points around the roots and the vertex:
- Let [tex]\( x = -5 \)[/tex]:
[tex]\[ y = (-5)^2 + 2(-5) - 8 = 25 - 10 - 8 = 7 \][/tex]
So the point is:
[tex]\[ (-5, 7) \][/tex]
- Let [tex]\( x = 5 \)[/tex]:
[tex]\[ y = 5^2 + 2 \cdot 5 - 8 = 25 + 10 - 8 = 27 \][/tex]
So the point is:
[tex]\[ (5, 27) \][/tex]
4. Plotting the Points:
From our calculations, we have the following points to plot on the graph:
- Roots or x-intercepts: [tex]\( (2, 0) \)[/tex] and [tex]\( (-4, 0) \)[/tex]
- Vertex: [tex]\( (-1, -9) \)[/tex]
- Additional Points: [tex]\( (-5, 7) \)[/tex] and [tex]\( (5, 27) \)[/tex]
5. Drawing the Parabola:
- Plot the points on the coordinate plane.
- Draw a smooth curve passing through these points, forming the characteristic "U" shape of a parabola.
Remember that the parabola opens upwards since the coefficient of [tex]\( x^2 \)[/tex] (which is [tex]\( a = 1 \)[/tex]) is positive. This confirmation supports the shape formed by the plotted points.
By following these steps, you will see a comprehensive graph of the quadratic equation [tex]\( y = x^2 + 2x - 8 \)[/tex] on the accompanying set of axes.
1. Find the Roots (Zeros) of the Equation:
The roots are the values of [tex]\( x \)[/tex] for which [tex]\( y = 0 \)[/tex].
For the equation [tex]\( x^2 + 2x - 8 = 0 \)[/tex]:
The roots (or x-intercepts) are:
[tex]\[ x = 2 \quad \text{and} \quad x = -4 \][/tex]
So, the points (roots) are:
[tex]\[ (2, 0) \quad \text{and} \quad (-4, 0) \][/tex]
2. Find the Vertex of the Parabola:
The vertex of a quadratic equation [tex]\( y = ax^2 + bx + c \)[/tex] is given by:
[tex]\[ x = -\frac{b}{2a} \][/tex]
For [tex]\( y = x^2 + 2x - 8 \)[/tex], [tex]\( a = 1 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -8 \)[/tex]. Thus:
[tex]\[ x = -\frac{2}{2 \cdot 1} = -1 \][/tex]
Plugging [tex]\( x = -1 \)[/tex] back into the equation to get the corresponding [tex]\( y \)[/tex]-value:
[tex]\[ y = (-1)^2 + 2(-1) - 8 = 1 - 2 - 8 = -9 \][/tex]
Therefore, the vertex of the parabola is:
[tex]\[ (-1, -9) \][/tex]
3. Choose and Calculate Additional Points:
To have a well-formed graph, we select points around the roots and the vertex:
- Let [tex]\( x = -5 \)[/tex]:
[tex]\[ y = (-5)^2 + 2(-5) - 8 = 25 - 10 - 8 = 7 \][/tex]
So the point is:
[tex]\[ (-5, 7) \][/tex]
- Let [tex]\( x = 5 \)[/tex]:
[tex]\[ y = 5^2 + 2 \cdot 5 - 8 = 25 + 10 - 8 = 27 \][/tex]
So the point is:
[tex]\[ (5, 27) \][/tex]
4. Plotting the Points:
From our calculations, we have the following points to plot on the graph:
- Roots or x-intercepts: [tex]\( (2, 0) \)[/tex] and [tex]\( (-4, 0) \)[/tex]
- Vertex: [tex]\( (-1, -9) \)[/tex]
- Additional Points: [tex]\( (-5, 7) \)[/tex] and [tex]\( (5, 27) \)[/tex]
5. Drawing the Parabola:
- Plot the points on the coordinate plane.
- Draw a smooth curve passing through these points, forming the characteristic "U" shape of a parabola.
Remember that the parabola opens upwards since the coefficient of [tex]\( x^2 \)[/tex] (which is [tex]\( a = 1 \)[/tex]) is positive. This confirmation supports the shape formed by the plotted points.
By following these steps, you will see a comprehensive graph of the quadratic equation [tex]\( y = x^2 + 2x - 8 \)[/tex] on the accompanying set of axes.
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