How to write the quadratic function.

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Discussion Overview

The discussion centers on how to write a quadratic function with specific characteristics, namely a maximum value of 4 and an axis of symmetry at x = -2. The focus is on the vertex form of the quadratic equation and the implications of the parameters involved.

Discussion Character

  • Technical explanation, Mathematical reasoning

Main Points Raised

  • One participant asks for suggestions on writing a quadratic function with a maximum value of 4 and an axis of symmetry at x = -2.
  • Another participant notes that the vertex form of a quadratic can be expressed as $$y(x)=a(x-h)^2+k$$ and questions where the vertex should be, emphasizing that 'a' must be negative for the quadratic to have a maximum.
  • A subsequent reply reiterates the vertex form and confirms that 'a' should be negative, proposing the function $$y(x)=-a(x+2)^2+4$$ as a potential solution.
  • Another participant agrees with the proposed function, stating that as long as $0

Areas of Agreement / Disagreement

Participants generally agree on the form of the quadratic function and the conditions for 'a', but there are multiple expressions proposed, reflecting different perspectives on how to represent the function.

Contextual Notes

The discussion does not resolve the specific value of 'a' and leaves open the interpretation of the function based on the conditions provided.

mathlearn
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Write a quadratic function y whose maximum value is 4 and the axis of symmetry of the graph is x=-2.

Suggestions?
 
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Where should the vertex be? Recall the vertex form of a quadratic may be written as:

$$y(x)=a(x-h)^2+k$$

where the vertex is at $(h,k)$. Since the quadratic is to have a maximum, what can we say about $a$?
 
MarkFL said:
Where should the vertex be? Recall the vertex form of a quadratic may be written as:

$$y(x)=a(x-h)^2+k$$

where the vertex is at $(h,k)$. Since the quadratic is to have a maximum, what can we say about $a$?

Since the quadratic is to have a maximum , 'a' should be a negative.

$$y(x)=-a(x+2)^2+4$$

Correct?
 
mathlearn said:
Since the quadratic is to have a maximum , 'a' should be a negative.

$$y(x)=-a(x+2)^2+4$$

Correct?

Yes, as long as $0<a$, then you have correctly given the family of quadratics that meet the stated criteria. :D

Alternately, you could state:

$$y(x)=a(x+2)^2+4$$ where $a<0$.
 
Thank you very much :)
 

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