MHB How to write the quadratic function.

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To write a quadratic function with a maximum value of 4 and an axis of symmetry at x = -2, the vertex form of the function is used: y(x) = a(x - h)² + k, where the vertex is at (h, k). The vertex should be at (-2, 4) since the maximum occurs there. The coefficient 'a' must be negative to ensure the parabola opens downwards, indicating a maximum point. Thus, the function can be expressed as y(x) = -a(x + 2)² + 4, with a > 0. This formulation meets the specified criteria for the quadratic function.
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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.

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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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