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maximum and minimum of a quadratic function
- Thread starter littlebit
- Start date
would someone mind explaining how to find the maximum or minimum of
y=x^2-2x-48
All quadratics graph as parabolas. The parabola either opens up or down, depending on the sign of leading coefficient (the coefficient of the x^2 term).
The max or min value of the function is at the vertex of the parabola. The vertex is a minimum for parabolas that open upward (is U-shaped, having a positive leading coefficient).
The vertex gives the maximum value if the parabola opens downward. Make sense?
The x value of the vertex is “-b/2a”, where “a” is the leading coefficient and “b” is the coefficient of the x term. In your problem, a = 1 and b = -2.
After finding the x value of the vertex, plug that x value into the equation to find the y value. This value is either your maximum or your minimum.
midwood_trail
New member
- Joined
- Nov 9, 2008
- Messages
- 9
littlebit said:would someone mind explaining how to find the maximum or minimum of
y=x^2-2x-48
Since the coefficient of the x^2 term is positive, then this parabola will have minimum value.
This means the parabola will open upward.
In either maximum or minimum values, you need to find the vertex of the parabola.
Do you know how to find the vertex of a parabola written as it was given to you?