In quadratic function, why is the larger | a | the smaller the opening of parabola?

In quadratic function, why is the larger | a | the smaller the opening of parabola?

If the absolute value of a is larger, the tangent slope of parabola will change faster, which is reflected in the image that the opening is smaller
I don't know if the questioner can understand, but this statement is very convincing

For quadratic function, the larger the value of a, the smaller the opening of parabola. Why? Consider y = 10x ^ 2 + 3x + 1

The speed of increase or decrease increases, that is, the change of tangent slope increases

The larger | a | is, the smaller the opening of parabola is, and the smaller | a | is, the larger the opening of parabola is

In extreme terms, you can think of a quadratic function like y = ax ^ 2,
When a approaches 0, | a | is the smallest. At this time, the function is basically y = 0, which is basically the X axis, so the opening is large
When a approaches positive and negative infinity, | a | is the largest. At this time, the two branches of y = ±∞ parabola almost coincide, so the opening is small

In the plane rectangular coordinate system, the circle C with radius r intersects with points D (1,0), e (5,0) on the x-axis and is tangent to the positive half axis of y-axis on point B
Points a and B are symmetric about X axis. Point P (a, 0) moves on the positive half axis of X axis, making a straight line AP, making eh perpendicular to AP and H
(1) Find the coordinates of center C and the value of radius R
(2) The triangle POA and the triangle Phe vary with the motion of point P. if they are congruent, the value of a is obtained
(3) If a = 6, try to judge the position relationship between the straight line AP and the circle C

(1) if BC is connected, then BC ⊥ Y axis
Take the midpoint m of De and connect cm, then cm ⊥ X axis
∵OD=1,OE=5,∴OM=3.
∵ ob2 = OD · OE = 5, ∵ ob =. ∵ center C, radius r = 3
（2）∵△POA≌△PHE,∴PA=PE.
∵OA=OB= ,OE=5,OP=a,∴ ,
∴
(3) Solution 1:
Let the tangent at (t is the tangent point) of ⊙ C through point a intersect the positive semiaxis of X at Q. let Q (m, 0), then QE = m-5, QD = M-1,
QT=QA-AT=QA-AB=
From OT2 = OE · OD, we get
∵
∵ a = 6, point P (6,0) is on the right side of point Q, and the ⊙ line AP is away from ⊙ C

In the plane rectangular coordinate system, take the negative half axis of the x-axis as the starting edge of the angle, if the terminal edges of the angles α and β intersect at the points of the unit circle respectively
(12 | 13,5 | 13) and (- 3 | 5,4 | 5), then sin α cos β is equal to
A.-36|65
B.-3|13
C.4|13
D.48|65

The point (12 | 13,5 | 13) is in the first quadrant, and the starting edge is in the negative half axis, so the angle α is obtuse, the hypotenuse = root sign (12 | 13) ^ 2 + (5 | 13) ^ 2 = 1, the value of sin α is negative, sin α = - sin (π - α) = - (5 / 13) / 1 = - 5 / 13,
The point (- 3 | 5,4 | 5) is in the second quadrant, so β is an acute angle, cos β is a positive value, hypotenuse = root sign (3 | 5) ^ 2 + (4 | 5) ^ 2 = 1, cos β = (3 / 5) / 1 = 3 / 5,
Sin α cos β = - 5 / 13 * 3 / 5 = - 3 / 13, so choose B

In the plane rectangular coordinate system, there are two straight lines, y = 3 / 5x + 9 / 5 and y = - 3 / 2x + 6. Their intersection points are p and a and B with X axis respectively
(1) Finding the coordinates of a, B and P
(2) Finding the area of △ PAB

1. Straight line: y = (3 / 5) x + 9 / 5, let y = 0, find out x = - 3, this straight line intersects with X axis at a (- 3,0); straight line y = (- 3 / 2) x + 6, let y = 0, find out x = 4, this straight line intersects with y axis at B (4,0); the equations of two straight lines are established to form a system of binary linear equations: y = (3 / 5) x + 9 / 5, y = (- 3 / 2) x + 6

In the plane rectangular coordinate system, there are two straight lines: y = x + 3 and y = - 2x + 6. Their intersection points are p, and their intersection points with the X axis are A.B 1

If y = x + 3 and y = - 2x + 6, the solution is x = 1, y = 4, so p (1,4);
In y = x + 3, let y = 0 get x = - 3, so a (- 3,0);
In y = - 2x + 6, let y = 0 get x = 3, so B (3,0);
So the area of triangle ABP is 1 / 2 * 6 * 4 = 12

In the plane rectangular coordinate system, if the line y = 1 / 2x + 2 moves up two units to get the line m, then the intersection coordinate of the line m and the X axis is

If the line m is y = 1 / 2x + 4, then the intersection coordinate of the line m and the X axis is (- 8,0)

In the plane rectangular coordinate system, if the intersection of the line y = x-m and the line y = 2X-4 is on the X axis, then M =?

m=2

In the plane rectangular coordinate system, there are two straight lines: y = 35x + 95 and y = - 32 + 6. Their intersection points are p, and their intersection points with X axis are a and B respectively. (1) find the coordinates of a, B and p; (2) find the area of △ PAB

(1) Let P (x, y), from the meaning of the question, we know that y = 35x + 95y = − 32x + 6, ∧ x = 2Y = 3, ∧ P (2, 3). The coordinates of the intersection a of the line y = 35x + 95 and the x-axis are (- 3, 0), and the coordinates of the intersection B of the line y = - 32x + 6 and the x-axis are (4, 0)