Axis of Symmetry of y=a(x-r_1)(x+r_2)

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The equation of the axis of symmetry for the parabola represented by y=a(x-r_1)(x+r_2) can be derived using the vertex formula x=-b/2a. In this case, b is identified as -a(r_1 - r_2) and a remains as the coefficient of x^2. The vertex's x-coordinate, which is the axis of symmetry, is thus calculated as x=(r_1 + r_2)/2. The discussion emphasizes understanding how to extract the axis of symmetry and vertex from the given quadratic equation. Mastery of these concepts is essential for analyzing parabolic functions effectively.
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Problem 1.
Consider the equation y=a(x-r_1)(x+r_2)
Write the equation of the axis of symmetry.
 
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That is just a simple parabola:
y = ax^2 - ax(r_1 - r_2) - ar_1r_2
Where is the 'vertex' of this parabola? What is its X coordinate?
 
How do you get the axis of symmetry..vertex..etc...

How would you get the axis of symmetry: x=-b/2a from b=-ax(r_1-r_2) and a= 1 or just a?
 
How do you get the axis of symmetry..vertex..etc...

How would you get the axis of symmetry: x=-b/2a from b=-ax(r_1-r_2) and a= 1 or just a?
 
How do you get the axis of symmetry..vertex..etc...

How would you get the axis of symmetry: x=-b/2a from b=-ax(r_1-r_2) and a= 1 or just a?
 
How do you get the axis of symmetry..vertex..etc...

How would you get the axis of symmetry: x=-b/2a from b=-ax(r_1-r_2) and a= 1 or just a?
 
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