How to write the quadratic function.

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SUMMARY

The discussion focuses on constructing a quadratic function with a maximum value of 4 and an axis of symmetry at x = -2. The vertex form of the quadratic function is given as y(x) = a(x - h)² + k, where the vertex is at (h, k). For the specified conditions, 'a' must be negative to ensure the function opens downward, leading to the function y(x) = -a(x + 2)² + 4, where a > 0. An alternative representation is y(x) = a(x + 2)² + 4, with the condition that a < 0.

PREREQUISITES
  • Understanding of quadratic functions and their properties
  • Familiarity with vertex form of quadratic equations
  • Knowledge of the significance of the vertex in determining maximum or minimum values
  • Basic algebra skills for manipulating equations
NEXT STEPS
  • Study the properties of quadratic functions, focusing on vertex form
  • Learn how to determine the maximum and minimum values of quadratic functions
  • Explore the effects of varying the coefficient 'a' in the vertex form
  • Practice writing quadratic functions given specific conditions for maximum or minimum values
USEFUL FOR

Students learning algebra, educators teaching quadratic functions, and anyone interested in mastering the vertex form of quadratic equations.

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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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