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 :)
 

Thread 'Area of Overlapping Squares'

Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?
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