It is known that the parabola y = (- 1 / 7) x ^ 2 + BX + C and the positive half axis of X axis intersect at two points a and B, ab = 4, P is a point on the parabola, and its abscissa is 9, Angle PbO = 135 degrees, cot angle PAB = 7 / 3 (1) Finding the coordinates of point P (2) finding the relation of parabola It's a problem about quadratic function

It is known that the parabola y = (- 1 / 7) x ^ 2 + BX + C and the positive half axis of X axis intersect at two points a and B, ab = 4, P is a point on the parabola, and its abscissa is 9, Angle PbO = 135 degrees, cot angle PAB = 7 / 3 (1) Finding the coordinates of point P (2) finding the relation of parabola It's a problem about quadratic function

1. Drawing. P (9, y)
y/(4+y)=7/3,y=-7
p(9,-7)
2. P point substituting: - 7 = - 81 / 7 + 9b + C
(-1/7)x^2+bx+c=0
AB=√[(x1+x2)^2-4x1x2]=√[(7b)^2+28c=4
b=,c=

Given that the parabola y = ax & # 178; + BX + C passes through three points a (0,1), B (1,3) and C (- 1,1), then a =? B =? C =?

Three point substitution
c=1
a＋b＋c=3
a－b＋c=1
The solution is a = 1, B = 1, C = 1

If the parabola y = ax & # 178; BX + C passes through three points a (- 1,0) B (3,0) C (0,1), then a =, B =, C=

Bring in:
a-b+c=0；
9a+3b+c=0；
c=1；
a-b=-1；
9a+3b=-1；
12a=-4；
a=-1/3;
b=2/3;
If you ask questions from your mobile phone, just click "satisfied" in the upper right corner of the client

The tangent equation of curve y = X3 + X + 1 at point (1,3) is ()
A. 4x-y-1=0B. 4x+y-1=0C. 4x-y+1=0D. 4x+y+1=0

∵ y = X3 + X + 1, ∵ y ′ = 3x2 + 1, let x = 1 get tangent slope 4, ∵ tangent equation is Y-3 = 4 (x-1), that is 4x-y-1 = 0, so select a

Given that the area of the triangle formed by the tangent of the curve y = X3 at the point (a, A3) (a is not equal to 0) and the line x = a is 1 / 6, find the value of A

The tangent slope is 3A (2), and the tangent equation is y = 3A (2) x-2a (3),
The height of the triangle is a (3), the bottom is x = 2A / 3 of the tangent when y = 0, and the area is a (4) / 3
The value of a is a positive or negative quarter root