How to draw the function image of grade two? I know the general steps of drawing a function image, that is, I don't know when the points on the image should be connected into line segments, or straight lines or rays, or even a series of points, and what to do if the value of the independent variable is infinite? Please answer in detail one by one and tell me the knowledge about the content of the image

Function image (y = ax + B; we discuss when a ≠ 0); whether it is 1, line segment; 2, ray; 3, straight line; 4, a string of points?; is determined according to the value range of variable X:
1: The value range of line segment x is: between two numerical values (such as: 1 ≤ x ≤ 9) --- analogy -- drawing key: two end points
2. The value range of ray x is: one side of a certain value on the number axis (such as: 1 ≤ X or X ≥ 1) --- analogy -- drawing key: endpoint, any point
3. The value range of straight line X is: any number on the number axis (that is, the value you said is infinite). Key to drawing: any two points
4. The value range of a string of points x is: several given values (such as: x = - 1; 2; 3; etc.) key: given value points
Of course: but when a = 0, the image is a straight line parallel (B ≠ 0) or coincident (b = 0) with the X axis
I hope I can help you solve the problem,

In the plane rectangular coordinate system, if a < 0 and b > 0 are known, then the point P (a, b) must be in the fourth quadrant?

The abscissa is negative
The ordinate is a positive number
Beta Quadrant
In the plane rectangular coordinate system, if a < 0 and b > 0 are known, then the point P (a, b) must be in the second quadrant

Given the first-order function y = [2m + 4] x + [3-N], when m = - 1, n = 2, find the area of the triangle formed by the image of the first-order function and the two coordinate axes

When m = - 1, n = 2, y = 2x + 1,
So if the intersection point a with X axis, then when y = 0, 2x + 1 = 0, x = - 1 / 2, OA = 1 / 2
When x = 0, y = 1, OB = 1
So the s triangle AOB = 1x1 / 2 * 1 / 2 = 1 / 4

If point P (M + 1, m) is known, then point P cannot be in the () quadrant
A. Four B. three C. two D. one

The abscissa of the easy to get point P is larger than its ordinate. In the four quadrants, only the abscissa of the point in the second quadrant is always smaller than its ordinate, so C

If the image of the first-order function y = (m-1) x + m + 2 does not pass through the third quadrant, then the range of M_________

The answer upstairs is too hasty
Why can't m be equal to 1?
On the one hand, this kind of problems should be combined with numbers and shapes, and we must also pay attention to the fact that the slope does not exist and the slope is zero, which is often the trap of the problem!
The positive solution is - 2

Given that the coordinates of point P are (m-1,2-m), which quadrant is point P impossible to be in?

Quadrant 1
m-1>0 and 2-m>0
m>1 and m

If a linear function y = - x square + 2 (M + 2) x + m Square-1 has a maximum value of 9, find M

The quadratic function y = - X & # 178; + 2 (M + 2) x + M & # 178; - 1 has a maximum value of 9,
Formula: y = - [x & # 178; - 2 (M + 2) x + (M + 2) & # 178;] + (M + 2) & # 178; + M & # 178; - 1
=-[x-(m+2)]²+(m+2)²+m²-1
When x = (M + 2), y gets the maximum value (M + 2) &# 178; + M & # 178; - 1 = 9
∴2m²+4m-6=0
∴m²+2m-3=0
‖ M = - 3 or M = 1

If P [a, A-1] is known, then p cannot be in several quadrants

Second quadrant
Because when a < o,
A-1 must be less than zero, not in the second quadrant

If the image of the first-order function y = ax + 1 + a squared intersects with the y-axis at the point (0,5), and the image passes through the first, second and third quadrants, then the function relation is expressed

From the function passing through the point (0,5), we can get 1 + a = 5 and get a = 4, so y = 4x + 5

Given that 2 / X of X + x = 4, which quadrant is the midpoint (2 / X of X + X, 2 / X of x-x) in the plane rectangular coordinate system?

x+2/x=4
(x+2/x)²=16
x²+4+(2/x)²=16
x²+(2/x)²=12
x²-4+(2/x)²=12-4
(x-2/x)²=8
x-2/x=±2√2
Maybe in the first quadrant, or in the fourth quadrant