If the parabola y x x b passes through the points a,1 4 and a,y1 , then the value of y1 is ? 35

The solution y=x +x+b can be transformed ;

x+1 2) +b2-1 4 is obvious when this formula comes out;

Because y=(x+1 2) +b2-1 4>= -1 4+b >=-1 4 because the point (a, -1 4) means b=0 and this point is the lowest point of the parabola, i.e., a=-1 2 , substituting -a i.e., 1 2 into the original equation gives y1=(1 2) +1 2=3 4

Substitute the values of the two points.

1/4=a^2 + a + b^2

y1 =a^2 -a + b^2

y1=1/4-2a

(1) a(3, 0), b(0, 3) parabola through a, c (both on x-axis): y = a(x - 3)(x - 1) x = 0, y = 3a = 3, a = 1 y = (x - 3)(x - 1) = x2 - 4x + 3 (2) oa = ob = 3, abo is an isosceles right triangle abo is similar to adp dp ab, dp = slope of da ab = -1, slope of dp = 1, Analytic formula for dp: y = 1(x + 1) = x + 1 in conjunction with y = -x + 3, p(1, 2) (3) e(e, e2 - 4e + 3), 1< e < 3 Area of ade s1 = (1 2)(3 + 1)(-e2 + 4e - 3) = 2(-e2 + 4e - 3) Area of quadrilateral APCE s2 = area of ACP + area of ace = (1 2)(3 - 1)*2 + 1 2)*2*(-e2 + 4e - 3) = 2 -e2 + 4e - 3 s1 = s2 2(-e2 + 4e - 3) = 2 -e2 + 4e - 3 -e2 + 4e - 3 = 2 e2 - 4e + 5= 0 = 16 - 20 = -4 < 0 e does not exist.

1) Point a(1,a) on the parabola y=x2, a=1;

The coordinates of point A are: (1,1);

2) Suppose there is a point p, according to OAP refers to a fierce isosceles triangle, OA=OP or OA=AP or OP=AP, so the coordinates of point P are: (

root number 2, disturbing friends 0); (root number 2, 0); (Acacia 2,0); （1，0）

The correct number is 4.

Reason: y=ax 2+bx+c(a<0) is pretended to be dotted (-1,0), and satisfies 4a+2b+c>0,0=a-b+c=0,b=a+c, there are 4a+2(a+c)+c>0, that is, 2a+c>0, (a<0, then c>0,)

2a+c>0, a+c>0 established.

Crack stove 2a+c>0, c>-2a, 4a+2b+c>0, 4a+2b-2a>0 is established, that is, a+b>0 is established.

b=a+c,a+b+c=-a+a+c+c=2c>0 is established.

b = a+c, b 2-2ac-5a 2=(a+c) 2-2ac-5a 2=c 2-4a 2, and c>-2a>0, both sides squared, c 2>4a, c 2-4a>0 is true, that is, b 2-2ac-5a 2=(a+c) 2-2ac-5a 2=c 2-4a 2> the difference is 0.

The parabola y=x +x+b passes through the points (a,-1 4) and (-a,y1), 1 4=a 2+a+b 2, y1=a 2-a+b 2, and subtracts y1+1 4=-2a.

Lack of conditions to find 3 4

x 2+x+b 2>=b 2-1 4, so if and only if b = 0, the parabola passes through a point with an ordinate of -1 4. That is, the parabola is y=x 2+x; a=-1/2。Bring it in to get y1=3 4

The solution y=x +x+b can be transformed ;

x+1 2) +b2-1 4 This omission is obvious when it comes out;

Because y=(x+1 2) +b2-1 4>= 1 4+b >1 4 because the passing point (a, -1 4) means b=0 and this point is the lowest point of the parabola i.e., a=-1 Huiqinhu 2 , substituting -a i.e. 1 2 into the original equation gives y1=(1 2) +1 2=3 4

If you can't do it this way, we can transform the first scum from their independent variables and the range of values that can also solve the problem.

Solution: The parabola passes through the point a(4,0), 42+4b=0,b=-2, the analytic formula of the parabola is: y= x2-2x= (x-2)2-2, the axis of symmetry of the parabola is x=2, the point c(1,-3), the symmetry point c with respect to x=2 c (3,-3), the intersection point of the line ac and x=2 is d, because any point d can form an adc and in a triangle, the difference between the two sides is less than the third side, i.e. |ad-cd|The maximum value of the AC slip is taken when D is the point on the AC extension linead-cd|=ac substitute the coordinates of a and c two points, and get the analytic formula of the straight line through ac to brighten Bila;

Let the analytic formula of the line ac be y=kx+b, and the solution is: , the analytic formula of the line ac is y=3x-12, when x=2, y=-6, and the coordinates of point d are (2, -6).

So the answer is: (2,-6).

Substituting (1,2) and (1,4) into the parabolic equation y=ax +bx+c yields:

a+b+c=2, a-b+c=4

The sum of the two formulas yields: a+c=3