If the image of quadratic function y = kx2-6x + 3 intersects with X axis, then the value range of K is () A. K < 3B. K < 3 and K ≠ 0C. K ≤ 3D. K ≤ 3 and K ≠ 0

If the image of quadratic function y = kx2-6x + 3 intersects with X axis, then the value range of K is () A. K < 3B. K < 3 and K ≠ 0C. K ≤ 3D. K ≤ 3 and K ≠ 0

∵ the image of quadratic function y = kx2-6x + 3 has intersection with the X axis, and the equation kx2-6x + 3 = 0 (K ≠ 0) has real roots, that is △ = 36-12k ≥ 0, K ≤ 3. Because it is a quadratic function, K ≠ 0, then the value range of K is k ≤ 3 and K ≠ 0

Find the intersection coordinates of the image and X axis of the function y = x square - 5 / 6x + 1 / 6, and make sketch verification

X & # 178; - 5 / 6 x + 1 / 6 = 0 is the intersection of X axis
6x²-5x+1=0
(2x-1)(3x-1)=0
x1=1/2 ;x2=1/3

There are two different intersection points between the image of quadratic function y = KX square-6x-7 and X axis. How to find the value range of K?

The equation KX ^ 2-6x-7 = 0 has two different real roots
That is △ > 0
^2 is the square

The number of intersections between the image and the x-axis of the function y = ax ^ 3 - (a + 2) x ^ 2 + 6x-3 (a + 2)

When a = 0, a / 2 is meaningless

If we know the intersection point P (1, y) of the functions y = x + B and y = ax + 3, then what is the solution set of X + b > ax + 3?

b+1=a+3
b=a+2
x+b>ax+3
x+a+2>ax+3
The image intersection P (1, y) of the functions y = x + B and y = ax + 3,
So a is not equal to 1
(1-a)x>1-a
Discuss two situations
(1)
a0
x>1
(2)
If a > 1, then 1-A

If the intersection coordinates of the image of the function y = 3x + A and y = − 13X + B are (- 3,4), then a=______ ，b=______ ．

According to the meaning of the question, 3 × (- 3) + a = 4, - 13 × (- 3) + B = 4, the solution is a = 4 + 9 = 13, B = 4-1 = 3

If the image of function y = - 4 / 3x-4 intersects X axis with a and Y axis with B, then the coordinate of point a is the coordinate of point B?

Let x = 0, y = - 4 / 3 * 0-4 = - 4, B (0,4)
Let y = 0, x = (0 + 4) / (- 4 / 3) = - 3, a (- 3,0)

The number of intersections between the image of the function y = SiNx on the interval (0,2 π) and the X axis is

When x = π, y = 0

The function y = 1 + SiNx, X belongs to [0,2 π], and the number of intersections with y = 2,

y=1+sinx
y=2
Simultaneous equations
2=1+sinx
1=sinx
That is, x = π / 2 + 2K π
And because x is between [0,2 π]
So x = π / 2, the number of intersections is one

The number of intersections between the image of y = 1 + Sin & nbsp; X, X ∈ [0, 2 π] and the line y = 2 is ()
A. 0B. 1C. 2D. 3

If we make the image of y = 1 + Sin & nbsp; X on [0, 2 π], we can see that there is only one intersection