## Given x < 5 / 4, find the maximum value of function y = 4x-2 + 1 / (4x-5) Given x > 0, Y > 0 and 1 / x + 9 / y = 1, find the minimum value of X + y

1) X < 5 / 4 = = = > 5-4x > 0y = 4x-2 + 1 / (4x-5) = 4x-5 + 1 / (4x-5) + 3 = - [5-4x + 1 / (5-4x)] + 35-4x + 1 / (5-4x) > = 2 √ (5-4x) * 1 / (5-4x2) = 2 (if and only if 5-4x = 1 / (5-4x) takes the equal sign) ymax = - 2 + 3 = 12) x + y = (x + y) * (1 / x + 9 / y) = 1 + 9 / y + Y / x = 10 + 9x / y + Y / x =

### Given that the axis of symmetry of the parabola y = AX2 + BX + C is 2 and passes through the point (3, 0), then the value of a + B + C () A. Equal to 0 B. Equal to 1 C. Equal to - 1 D. Not sure

∵ the axis of symmetry of parabola y = AX2 + BX + C is 2,

According to the symmetry of quadratic function, the symmetry point of point (3,0) is (1,0),

∵ when x = 1, y = a + B + C = 0,

The value of a + B + C is equal to 0

So choose a

### Given that the axis of symmetry of the parabola y = ax ^ 2 + BX + C is x = 2 and passes through points (1,4) and (5,0), the analytical formula of the parabola is

The axis of symmetry is x = 2

y=a(x-2) ²+ k

Put two points in

4=a+k

0=9a+k

So a = - 1 / 2, k = 9 / 2

y=-(x-2) ²/ 2+9/2

That is, y = - X ²/ 2+x+5/2

### Given that the parabola y = AX2 + BX + 2 passes through the point (3,2), then the axis of symmetry of the parabola is a straight line _

∵ parabola y = AX2 + BX + 2 passes through point (3, 2),

∴9a+3b+2=2，

∴b=-3a，

The axis of symmetry of the parabola is a straight line x = - B

2a=--3a

2a=3

2，

That is, x = 3

2．

So the answer is: x = 3

2．

### It is known that the axis of symmetry of the parabola y = ax + BX + C is a straight line x = 2, and passes through points (1,4) and (5,0). Find the function expression corresponding to this parabola

Y = ax + BX + C x = 2, then - B / 2A = 2 passes through point (1,4) and point (5,0) 4 = a + B + C 0 = 25A + 5B + C, and a = - 1 / 2, B = 2, C = 5 / 2 is obtained

### Given the parabola y = - x square + BX + C, find the expression of the parabola through points a (4,0) and B (1,3) (1); (2) Note that the axis of symmetry of the parabola is a straight line L Let the point P (m, n) on the parabola be in the fourth quadrant, the symmetry point of point P with respect to straight line L is e, and the symmetry point of point e with respect to y axis is F. if the area of quadrilateral oapf is 20, find the value of M, n

1. If the parabola passes through a and B, then:

-16+4b+c=0

-1+b+c=3

==>b=4,c=0

==>y=-x^2+4x

2. P (m, n), straight line L: x = 2

==>E(4-m,n),F(m-4,n)

==>Soapf=S△OFP+S△AOP

=1/2*(m-m+4)*|n|+1/2*4*|n|

=4|n|=20

==>n=-5

==>-m^2+4m=-5

==>m=5

P (5, - 5)