Function y = 3x + 2 image and y = 1 / X image intersection set

Function y = 3x + 2 image and y = 1 / X image intersection set

3x+2=1/x
3x^2+2x-1=0
(3x-1)(x+1)=0
X = 1 / 3 or x = - 1
Description method:
{x | x = 1 / 3 or x = - 1}
Enumeration method:
{1/3,-1}

The description method is used to represent the set of all the 耟 = points (x, y) on the image of the function y = 2x + 1

A straight line at an angle of arctan2 to the x-axis and passing through a point (- 0.5,0) on the x-axis

Given the function y = f (x), set a = {(x, y) y = f (x)}, B = {(x, y) | x = a, y ∈ r}, where a is a constant, then set a ∩ B
At most () elements of

Because a is a function f (x), there is only one y value corresponding to an X
B is a straight line perpendicular to the X axis, x = a,
So if a is a point in the domain of F (x), then a ∩ B has an intersection,
If a is not a point in the domain of F (x), then a ∩ B has no intersection

The quadratic function y = ax square + BX + C (a) is known

It can be seen from the fact that the image is not under the x-axis
Opening upward, a > 0,
When a + B + C is the function value of x = 1, the image is always not below the X axis, so when x = 1, y ≥ 0
And ∵ a ∵ B ∵ B-A ∵ 0
∴(a+b+c)/(b-a)≥0
If M is less than 0, the formula holds

Suppose that the quadratic function f (x) has f (x) ≤ f (1 / 2) = 25 for all real numbers x, its image has two intersections with X axis, and the cubic sum of abscissa of these two points is 19, then the analytic expression of F (x) is?

Let me have a try. From the problem, y = f (x) image symmetry axis X = 1 / 2, let two intersection points a (x1.0) B (x2,0) X1 + x2 = 2 * 1 / 2 = 1 and X1 & # 179; + x2 & # 179; = 19 = (X1 + x2) [(x1 + x2) &# 178; - 3x1x2] 1-3x1x2 = 19, x1x2 = - 6X1 + x2 = 1, solve X1 = 3, X2 = - 2, let the analytic formula be f (x) = a (x-3) (x + 2) substituted into f (1

The function y = ax2-2x has and only has two points on the image whose distance from X axis is equal to 1, then the value range of a is______ ．

When a = 0, there are only two points on the function y = - 2x image, and the distance from the X axis is equal to 1, which satisfies the condition. When a > 0, the minimum value of the function is − 44a ＞− 1, that is, a > 1. When a < 0, the maximum value of the function is − 44a ＜ 1, that is, a ＜ - 1. In conclusion, the value range of a is a > 1 or a = 0 or a ＜ - 1, so the answer is: a > 1 or a = 0 or a ＜ - 1

High school quadratic function,
There is at least one intersection point between the image of quadratic function y = MX ^ 2 + (M-3) x + 1 and X axis on the right side of the origin, and the value range of M is calculated,

Use the root formula to find the value of the left point (that is, the root with negative sign before the root sign) and let it be less than or equal to 0

Quadratic function (high school)
For the quadratic function y = - x ^ 2-4x + 1, the following statement is true____ ?
a. The graph of a function is a parabola with its opening up
b. When x = - 2, the maximum value of the function is 3
c. When x ≤ - 2, the value of function y increases with the increase of X
d. The function image has no focus on the X axis

C. A should have the opening downward, B should have the maximum value 5, D should have the intersection with X axis

1. Given K ＜ - 4, then the minimum value of function y = 2x ^ 2-1 + K (x-1) when - 1 ≤ x ≤ 1 is____
2. Given that the function f (x) = - x ^ 2 + 2aX + 1-A has a maximum value of 2 when 0 ≤ x ≤ 1, then a=____
3. Given the function f (x) = x ^ 2-2x + 3, when x ∈ [T, t + 1], then the maximum value of the function is____
4. Given that the inequality x ^ 2 + 2x-k > 0 holds in [1, + ∞), find the value range of K
5. Given that the inequality x ^ 2 + 2kx-4 > 0 holds in [1,2], find the value range of K

1. The axis of symmetry of the function is x = - 0.5K, because K ＜ - 4, so the axis of symmetry is x ＞ 2, so the function decreases monotonically in [- 1,1], so the minimum value is f (1) = 12. It is easy to know that the axis of symmetry is x = - A. the scope of the axis of symmetry is discussed as follows: 1. When - a ≥ 1, that is, a ≤ - 1, the function decreases on [0,1], so the maximum value is f (0), Let f (0) = 2

The maximum value of quadratic function
F (x) = x square + | X-2 | - 1, X ∈ R, find the minimum value of F (x)
Silent --, that what, broken moon wing. I couldn't understand.

Classified discussion
When x > 2
F（X）=x^2+x-3=(x+1/2)^2-13/4
F（X）min=(2+1/2)^2-13/4=3
When x