Yes
When the quadratic is written in the form: y = ax2 + bx + c then if a > 0 y has a minimum if a < 0 y has a maximum and if a = 0 y is not a quadratic but y = bx + c, and it is linear. The maximum or minimum is at x = -b/(2a)
If x2 is negative it will have a maximum value If x2 is positive it will have a minimum value
maximum and minimum are both (-b/2a , c - (b^2/4a))
Find the maximum and minimum values that the function can take over all the values in the domain for the input. The range is the maximum minus the minimum.
The minimum is the vertex which in this case is 0,0 or the origin. There isn't a maximum.....
In theory you can go down the differentiation route but because it is a quadratic, there is a simpler solution. The general form of a quadratic equation is y = ax2 + bx + c If a > 0 then the quadratic has a minimum If a < 0 then the quadratic has a maximum [and if a = 0 it is not a quadratic!] The maximum or minimum is attained when x = -b/2a and you evaluate y = ax2 + bx + c at this value of x to find the maximum or minimum value of the quadratic.
It can't - unless you analyze the function restricted to a certain interval.
vertex
Yes
Standard notation for a quadratic function: y= ax2 + bx + c which forms a parabola, a is positive , minimum value (parabola opens upwards on an x-y graph) a is negative, maximum value (parabola opens downward) See related link.
When the quadratic is written in the form: y = ax2 + bx + c then if a > 0 y has a minimum if a < 0 y has a maximum and if a = 0 y is not a quadratic but y = bx + c, and it is linear. The maximum or minimum is at x = -b/(2a)
The vertex.
If x2 is negative it will have a maximum value If x2 is positive it will have a minimum value
Addition is the maximum or minimum function in math.
maximum and minimum are both (-b/2a , c - (b^2/4a))
It if the max or minimum value.