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The graph of the known quadratic function y = AX2 + BX + C (a ≠ 0) is shown in the figure ① ABC > 0; ② B < A + C; ③ 4A + 2B + C > 0; ④ 2C < 3B; ⑤ a + b > m (am + b) (m ≠ 1 real number) The correct conclusion is () A. 2 B. Three C. Four D. Five
If the opening is downward, a < 0; if the symmetry axis is on the right side of the Y axis, if a and B are different signs, then b > 0; if the intersection point of parabola and Y axis is above the X axis, C > 0, then ABC < 0, so ① is incorrect;
When x = - 1, the image is below the X axis, then y = A-B + C < 0, that is, a + C < B, so ② is incorrect;
If the symmetry axis is a straight line x = 1, then x = 2, the image is above the X axis, then y = 4A + 2B + C > 0, so ③ is correct;
x=-b
If 2A = 1, then a = - 1
If A-B + C < 0, then - 1
2b-b + C < 0, 2C < 3b, so ④ is correct;
When the opening is downward, when x = 1, y has the maximum value a + B + C; when x = m (m ≠ 1), y = AM2 + BM + C, then a + B + C > AM2 + BM + C, i.e. a + b > m (am + b) (m ≠ 1), so ⑤ is correct
Therefore, B
As shown in the figure, the vertex of the image with the quadratic function y=x2-2x-1 is known to be A. the image with the quadratic function y=ax2+bx and the X axis intersect the origin O and the other point C, and its vertex B is on the symmetry axis of the image with the function y=x2-2x-1. The vertex of the image with the quadratic function O and the X axis intersect the origin C and the other point C (1) Find the coordinates of point a and point C; (2) When the quadrilateral aobc is rhombic, the relation of function y = AX2 + BX is obtained
(1) ∵ y = x2-2x-1 = (x-1) 2-2,
As shown in the figure, the vertex of the image with the quadratic function y=x2-2x-1 is known to be A. the image with the quadratic function y=ax2+bx and the X axis intersect the origin O and the other point C, and its vertex B is on the symmetry axis of the image with the function y=x2-2x-1. The vertex of the image with the quadratic function O and the X axis intersect the origin C and the other point C (1) Find the coordinates of point a and point C; (2) When the quadrilateral aobc is rhombic, the relation of function y = AX2 + BX is obtained
（1）∵y=x2-2x-1=（x-1）2-2，
The coordinate of - 2 is the vertex of A
∵ the image of the quadratic function y = AX2 + BX intersects the origin O and the other point C, and its vertex B is on the symmetry axis of the image of the function y = x2-2x-1
The symmetry axis of the quadratic function y = AX2 + BX is a straight line x = 1,
The point C and the point o are symmetric about the line x = 1,
The coordinates of point C are (2, 0)
(2) Because the quadrilateral aobc is rhombic, points B and a are symmetric about the line OC,
Therefore, the coordinates of point B are (1, 2)
Because the image of the quadratic function y = AX2 + BX passes through points B (1,2), C (2,0),
therefore
a+b＝2
4a+2b＝0 ，
The solution
a＝−2
b＝4 ，
So the relation of the quadratic function y = AX2 + BX is y = - 2x2 + 4x
As shown in the figure, the vertex of the image with the quadratic function y=x2-2x-1 is known to be A. the image with the quadratic function y=ax2+bx and the X axis intersect the origin O and the other point C, and its vertex B is on the symmetry axis of the image with the function y=x2-2x-1. The vertex of the image with the quadratic function O and the X axis intersect the origin C and the other point C (1) Find the coordinates of point a and point C; (2) When the quadrilateral aobc is rhombic, the relation of function y = AX2 + BX is obtained
（1）∵y=x2-2x-1=（x-1）2-2，
The coordinate of - 2 is the vertex of A
∵ the image of the quadratic function y = AX2 + BX intersects the origin O and the other point C, and its vertex B is on the symmetry axis of the image of the function y = x2-2x-1
The symmetry axis of the quadratic function y = AX2 + BX is a straight line x = 1,
The point C and the point o are symmetric about the line x = 1,
The coordinates of point C are (2, 0)
(2) Because the quadrilateral aobc is rhombic, points B and a are symmetric about the line OC,
Therefore, the coordinates of point B are (1, 2)
Because the image of the quadratic function y = AX2 + BX passes through points B (1,2), C (2,0),
therefore
a+b＝2
4a+2b＝0 ，
The solution
a＝−2
b＝4 ，
So the relation of the quadratic function y = AX2 + BX is y = - 2x2 + 4x
As shown in the figure AF is the diameter of ⊙ o, the ⊙ C with OA as the diameter intersects with the chord ab of ⊙ o at point D, de ⊥ ob, and the perpendicular foot is e (1) D is the midpoint of ab; (2) De is the tangent of ⊙ C; （3）BE•BF=2AD•ED．
It is proved that: (1) connecting OD, ∵ OA is the diameter of ⊙ C,
It is known that AF is the diameter of circle O, and circle C with OA diameter intersects with chord ab of circle O at point D, de ⊥ ob
prove:
With OD, CD,
Because Ao is the diameter
So ∠ ADO = 90
And because AB is a string
So ad = dB (vertical diameter theorem)
AC = Co
So CD is the median of △ AOB
So CD ‖ ob
Because ∠ deo = 90
So ∠ CDE = ∠ deo = 90
So De is the tangent of the circle
As shown in the figure, the chord AB and radius OC of ⊙ o are extended and intersected at point D, BD = OA. If ∠ AOC = 105 °, then ∠ D=______ Degree
Connect ob,
∵BD=OA，OA=OB
So △ AOB and △ BOD are isosceles triangles,
Let ∠ d = x degree, then ∠ oba = 2x °,
Because ob = OA,
So ∠ a = 2x °,
In △ AOB, 2x + 2x + (105-x) = 180,
The solution is x = 25,
That is ∠ d = 25 °
It is known that in the circle O, OC is perpendicular to points AB and C, ab = 16, sin ∠ AOC = 3 / 5, (1) find the length and chord center distance of radius OA of circle O, (2) find cos ∠ AOC and Tan ∠ AOC
0
As shown in the figure, it is known that AB is the chord of ⊙ o, the radius OA = 20cm, ∠ AOB = 120 ° and the area of ⊙ AOB is obtained
The transition point O is OC ⊥ AB to C, as shown in the following figure: AOC = 12 ﹤ AOB = 60 °, AC = BC = 12ab, ﹤ in RT △ AOC, ﹤ a = 30 ° OC = 12oa = 10cm, AC = oa2 − oc2 = 202 − 102 = 103 (CM), ab = 2Ac = 203cm ﹥ the area of AOB = 12ab · OC = 12 × 203 × 10 = 1003 (cm2)
As shown in the figure, it is known that AB is the chord of ⊙ o, the radius OA = 20cm, ∠ AOB = 120 ° and the area of ⊙ AOB is obtained
The transition point O is OC ⊥ AB to C, as shown in the following figure: AOC = 12 ﹤ AOB = 60 °, AC = BC = 12ab, ﹤ in RT △ AOC, ﹤ a = 30 ° OC = 12oa = 10cm, AC = oa2 − oc2 = 202 − 102 = 103 (CM), ab = 2Ac = 203cm ﹥ the area of AOB = 12ab · OC = 12 × 203 × 10 = 1003 (cm2)
RELATED INFORMATIONS
- 1. As shown in the figure, it is known that AB is the chord of ⊙ o, the radius OA = 20cm, ∠ AOB = 120 ° and the area of ⊙ AOB is obtained
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