Maximum or Minimum Point of Quadratic Function

Some common questions that may pop out when talking about quadratic functions are about the maximum or minimum points, commonly called extreme point. Every quadratic equation has either a maximum point or a minimum point, depending on the value of a in the quadratic function y(x) = ax² + bx + c . Orimath Quadratic Solver have a built-in function capable of dealing with this task.

All you have to do is to check any check boxes with captions Maximum/Minimum ( see the picture above ). Then click as HTML to tell Orimath Quadratic Solver to solve the problem for you and tell you the steps required to get the answer.

Answer by Orimath Quadratic Solver :

Question Number 1 :
For this function y(x) = x² + 5x + 4 , answer the following questions :

A. Find the minimum/maximum point of the function !

Answer Number 1 :
The equation x² + 5x + 4 = 0 is already in ax²+bx+c=0 form.
By matching the constant position, we can derive that the value of a = 1, b = 5, c = 4.

1A . Find the minimum/maximum point of the function ! Back To Question
	Since the value of a = 1 is positive, the function y(x) = x² + 5x + 4 have a minimum point.
	Use the formula y'(x) = 0 , to find the value of x in the minimum point
	We have to find the function y'(x) first
	So we get y'(x) = 2x + 5 = 0
	Which means that 2x = -5
	Which means that x = -5/2
	So we get x = -2.5
	So the minimum point is ( x , y ) = ( -2.5 , y(-2.5) )
	Which is ( x , y ) = ( -2.5 , -2.25 )

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Problems related to Quadratic Equations that Orimath Quadratic Solver is capable to do :

Problems related to Quadratic Functions that Orimath Quadratic Solver is capable to do :

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