If the intersection of the image of the first order function y = (A-2) x + 2a-3 and the y-axis is above the x-axis, then the value range of a is I worked out a > 1.5, but the answer is a > 1.5 and a ≠ 2. Please explain a ≠ 2 in detail. Thank you

If the intersection of the image of the first order function y = (A-2) x + 2a-3 and the y-axis is above the x-axis, then the value range of a is I worked out a > 1.5, but the answer is a > 1.5 and a ≠ 2. Please explain a ≠ 2 in detail. Thank you

∵ the intersection point between the image of the first order function y = (A-2) x + 2a-3 and the y-axis is above the x-axis,
ν A-2 ≠ 0, 2a-3 ＞ 0 (coefficient of independent variable cannot be 0)
∴a≠2,a﹥1.5
If the intersection point between the image of the first order function y = (A-2) x + 2a-3 and the y-axis is above the x-axis, then the value range of a is: a > 1.5 and a ≠ 2

Given the first order function y = (3-K) x + (K-2) (k is a constant), the image passes through the first, second and third quadrants to find the value range of K To the most detailed process, write each step clearly From the meaning of the title 1.3-K>0 2.K-2>0 The solution is 2

The image passes through the first, second and third quadrants, so the image is inclined upward and does not pass through zero, so the slope is greater than zero, and the constant term is greater than zero, so 3-K > 0 K-2 > 0
I can't figure it out. I can't understand it

If the image of the primary function y= (2a-1) x+ (A-1) passes through the positive half axis of the Y axis and passes through the negative half axis of the X axis, the value range of a is () A. a＞1 Two B. a＞1 C. 1 2＜a＜1 D. a＜1 Two

According to the meaning of the question, the first order function image passes through the first, second and third quadrants,
So 2a-1 > 0 and A-1 > 0,
A > 1
Therefore, C

If the graph of the first order function y = KX + B (k is a constant and K ≠ 0) is shown in the figure, then the value range of X for Y > 0 is___ .

Because the coordinates of the intersection point of the line y = KX + B and the X axis is (- 2,0), it can be seen from the graph of the function that Y > 0 when x is less than - 2
Therefore, the value range of X for Y > 0 is: X ＜ - 2
So the answer is: X ＜ - 2

Given the function y = (2m-3) x + (m-1), if the intersection of its image and y-axis is on the negative half axis, what is the value range of M? Rt.y = (2m-3) x + (m-1), the intersection of its image and y-axis is on the negative half axis, and the value range of M is calculated

y=(2m-3)x+(m-1)
Intersection point P (0, Y1) with y axis
y1=m-1<0
M<1

Given that the image of the function y = x2 + 2 (a + 2) x + A2 has two intersections with the X axis, and both of them are on the negative half axis of the X axis, then the value range of a is______ .

Let the abscissa of the intersection point of parabola and X axis be x1, x2,
According to the meaning of the title, X1 + x2 = - 2 (a + 2) < 0
x1•x2=a2＞0 ②
And △ = b2-4ac > 0,
∴a+1＞0  ③
A ＞ - 1 and a ≠ 0
So fill in the blank answer: a ＞ - 1 and a ≠ 0

If there are two intersections between the image of function f (x) = x ^ 2-ax + 1 and the positive half axis of X axis, then the value range of a is?

The meaning of the title is as follows:
a>0,△=a^2-4>0
∴a>2

If the function f (x) = x ^ 2 + (K + 2) x + 2k-1 and the two intersections of the X axis are on the positive half axis of the x-axis, the value range of K should be calculated in detail, We use Veda's theorem. I can do it

Let f (x) = 0 two roots x1, x2
X1 + x2 = - (K + 2) > 0, K0 is obtained and k > 1 / 2 is obtained
So the range of K is K1 / 2

If the image of the function f (x) = x ^ 2 * lga-2a + 1 has two intersections with the X axis, then the value range of the real number a is

F (x) can be regarded as a quadratic function
Coefficient of quadratic term LGA
Coefficient of primary term 0
Constant term - 2A + 1
There are two intersections between the image and the x-axis
So the discriminant is > 0
0-4(-2a+1)(lga)>0
lga（a-1/2）>0
lga>0=>a>1
a-1/2>0=>a>1/2
So a > 1 or A0
To sum up, 0

Given that α is the third quadrant angle, try to determine the final edge position of 2 α

α is the third quadrant angle
Let α = (2k + 1) Π + X, K be an integer, 02 α = 2 * (2k + 1) Π + 2x
When 0, when x = Π / 4,2 α is on the axis between the first and second quadrants
When Π / 4
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