x^2=y+4>; =0
y & gt=-4
Generally f (x) = ax 2+bx+c (a is not zero).
=a(x^2+b/ax)+c
=a(x+b/(2a))^2+(4ac-b^2)/(4a)
When x=-b/(2a), f(x) takes the maximum value (a0).
(4ac-b^2)/(4a),2,y=ax^2+bx+c
=a(x^2+bx/a+c/a)
=a[(x+b/2a)^2+(4ac-b^2)/4a^2]
=a(x+b/2a)^2+(4ac-b^2)/4a
When a > 0, y has a maximum value, and only when x=-b/2a, the maximum value of y is: (4ac-b 2)/4a.
When a < 0, y has a minimum value, and only when x=-b/2a, the minimum value of y is: (4ac-b 2)/4a.
For example, if ..., 2, y = x 2 -4, the opening is upward, and the vertex (0,-4) is the lowest point of the function.
Then, the minimum value of y is -4, and there is no maximum value.
Formula one, find the vertex.
Generally speaking, y = ax 2+bx+c (a is not 0).
Formula: y = a (x+b/2a) 2+c-b 2/4a.
The vertex is (-b/2a, c-b 2/4a).
If, a>0, the opening is upward, then ymin = c-b 2/4a, and there is no maximum.
What if a ... 1, y=x? -4:x = 0, y=-4 has a minimum value.
y=a(x-b)? +c:a≠0
A > 0: x = b has a minimum value of y = C.
A < 0: X = B has the maximum value Y = C, 0. How to find the maximum and minimum of quadratic function in mathematical problems?
There is no basic inequality.
How to do it by matching method
For example, find y = x 2-4.
Literally, y is equal to the square of x minus 4.