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.