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Let y = x2-2x + 2 and y = - x2 + ax + B be perpendicular to each other 1. Find the relationship between a and B 2. If a is greater than 0 and B is greater than 0, find the maximum value of a times B.
For parabola: y = - x2 + 2x-3, the following conclusion is correct ()
A. There are two intersections with the x-axis B. the opening is upward C. the coordinates of the intersection with the y-axis are (0,3) D. the coordinates of the vertex are (1, - 2)
A. If ∵ △ = 22-4 × (- 1) × (- 3) = - 8 ＜ 0, there is no intersection point between the parabola and the x-axis, this option is wrong; B, if ∵ quadratic coefficient - 1 ＜ 0, the opening of the parabola is downward, this option is wrong; C, if x = 0, y = - 3, the coordinates of the intersection point between the parabola and the y-axis are (0, - 3), this option is wrong; D, ∵ y = - x2 + 2x-3 = - (x -...)
The image of the square of parabola y = x-2x + 1 has some relations with the intersection of x-axis
△=b²-4ac=4-4=0
So there's only one point of intersection with the x-axis
The coordinates of the intersection of the parabola y = - 3x ^ 2 + 2x + 1 and the X axis are____ , and y-axis____ The coordinates of the point of intersection are____ Because a = - 3, y has the most____ Value, when x=___
The coordinates of the intersection of the parabola y = - 3x ^ 2 + 2x + 1 and the X axis are____ The coordinates of the intersection of the Y-axis and the y-axis are____ Because a = - 3, y has the most____ Value, when x=____ The maximum value of the function y is 0____ The value is____ .
The coordinate of the intersection of the object line y = - 3x ^ 2 + 2x + 1 and the X axis is____ The coordinates of the intersection of the Y-axis and the y-axis are____ Because a = - 3, y has the most____ Value, when x=____ The maximum value of the function y is 0____ The value is____ When the ordinate is 0y = 0, the intersection of - 3x ^ 2 + 2x + 1 = 03x ^ 2-2x-1 = 0 (x-1) (3x + 1) = 0 x = 1 or x = - 1 / 3 with x-axis
Given the parabola y = x & sup2; - 2x-8, if the two intersections of the parabola and the X axis are AB and its vertex is p, the area of the triangle ABP is obtained
Let y = x & sup2; - 2x-8 = 0, calculate x = 4, - 2. A (4,0) B (- 2,0). The abscissa of the parabola vertex must be the key point of AB, XP = 1, and YP = - 9. Draw a picture, take AB as the bottom, the ordinate of P point is high, s = (1 / 2) * 6 * 9 = 27
If we know that the parabola y = (x-1) ^ 2-9 intersects the x-axis A and B, and the vertex P, we can find the area of triangle ABP
y=0,x1=4,x2=-2
x=1,y=-9
Then a (4,0), B (- 2,0), P (1,9)
Area = 1 / 2 [4 - (- 2)] * 9 = 27
The parabola y = xx-2x + K + 1 has two intersections with the X axis
Analysis
b²-4ac>0
therefore
4-4x1x(k+1)>0
4-4(k+1)>0
4>4(k+1)
k+1
Given that the parabola y = x2 + 2 (K + 1) x-k has two intersections with the X axis, and the two intersections are on both sides of the line x = 1, then the value range of K is______ ．
∵ parabola y = x2 + 2 (K + 1) x-k has two intersections with X axis, and the two intersections are on both sides of the straight line x = 1, ∵ when x = 1, y ＜ 0, so substituting x = 1 into the analytical formula, we get: 1 + 2 (k + 1) - K ＜ 0 ∵ K + 3 ＜ 0, we get k ＜ - 3; so the value range of K is k ＜ - 3
It is known that the parabola y = (k-1) x & # 178; + 2kx + K-2 has two different intersections with the x-axis. (1) the range of value of K is obtained. (2) when k is an integer and the solution of the equation 3x = kx-1 about X is satisfied, the expression of the parabola is obtained
(1) For the quadratic equation (k-1) x & # 178; + 2kx + K-2 = 0, the discriminant > 0 (2k) &# 178; - 4 (k-1) (K-2) > 03k-2 > 0k > 2 / 3. In conclusion, when k > 2 / 3 and K ≠ 1 (2) 3x = kx-1 (K-3) x = 1K = 3, the equation has no solution, K ≠ 3x = 1 / (K-3), and the solution is negative, 1 / (K-3)
It is known that the parabola y = x ^ 2-2 (k-1) x + K ^ 2 has an intersection with the X axis, and the value range of K is obtained
y = x² - 2（k - 1）x + k²
△ = 4（k-1）² - 4k²
= - 8k + 1
The result is - 8K + 1 ≥ 0
The solution is k ≤ 1 / 8
RELATED INFORMATIONS
- 1. Given that the parabola y = x2 + 2 (K + 1) x-k has two intersections with the X axis, and the two intersections are on both sides of the line x = 1, then the value range of K is______ ．
- 2. The proof of 〔∫ (a, b) f (x) DX 〕 & # 178 in double integral; ＝∫(a,b)f(x)dx∫(a,b)f(y)dy
- 3. We've learned about the composition of triangles. What are the elements
- 4. Both sides____ An isosceles triangle is called an isosceles triangle_______ This triangle is called________
- 5. A point that is equidistant from all three sides of a triangle is the center of the triangle
- 6. Help to answer Thank you, y = (x-1) cosx + 2011 / X for dy / DX | x = 1
- 7. For any real number x, y ∈ R, SiNx + cosy = 2Sin (X-Y / 2 π / 4) cos (x y / 2 - π / 4), then sin13 π / 24. Cos5 π / 24 = =? Which formula is used in this problem? I don't want it. I need the formula
- 8. The proof of dy / DX f（x）＝cos x＋sin x／cos x－sin x It is proved that f '(x) = 2 / 1-sin 2x . .
- 9. Try to derive from DX / dy = 1 / y ': D ^ 2x / dy ^ 2 = - y' '/ (y') ^ 3 the problem of D [1 / y '] / DX} * [DX / dy] D [1 / y '] / DX} * [DX / dy] (because y' in 1 / y 'is the derivative of the function y = f (x), it is the function of X, so 1 / y' is also the function of X. if the function of X wants to derive y now, it needs the derivative method of composite function. For 1 / y ', it needs to derive x first, and then y) Since y is a function of X, why does 1 / y 'need to derive x from y after deriving x? Is it because the inverse function x is also a function of Y A little dizzy
- 10. The solution dy / DX + 1 / 3Y = 1 / 3 * (1-2x) y ^ 4 I feel the answer is wrong,
- 11. Given the function y = y1-y2, and Y1 is inversely proportional to x + 1, Y2 is positively proportional to X2, and x = - 2 and x = 1, the value of Y is 1. Find the functional relation of Y with respect to X
- 12. The linear equation which passes through the point (- 1, - 2) and is tangent to the curve y = 2x-x ^ 3 is solved by derivative
- 13. If the length of the line segment of the parabola y = - x squared + 4x-m on the x-axis is 4, then the value of M is zero
- 14. Proof: if there is a circle whose any two strings are equal, then the arcs of the two strings are equal
- 15. Parabola y = - x ^ 2 + (m-2) x + 3 (m-1) passing through point (3,0) Find the vertex coordinates of this parabola?
- 16. If the two ends of the line segment AB of length 2 slide on the parabola y2 = x, the minimum distance from the midpoint m of the line segment AB to the Y axis is 0______ ．
- 17. From a point P on the parabola y2 = 4x, the vertical foot of the parabola is m, and | PM | = 5. Let the focus of the parabola be f, then the area of △ MPF is______ ．
- 18. A straight line passing through the focus of Y (2) = 2px (P > 0) intersects with this parabola. The ordinates of the two intersections are Y1 and Y2 respectively
- 19. If the focus of y2 = 2x is f and the straight line passing through f intersects the parabola at two points AB, then the minimum value of | AF | + 4 | ab | Don't use the focal radius
- 20. As shown in the figure, it is known that the parabola Y1 = ax & # 178; + BX + C and the parabola y2 = x & # 178; + 6x + 5 are symmetric about the y-axis, and intersect with the y-axis at point m, and intersect with the x-axis at two points a and B If the image of a function y = KX + B passes through a point m and intersects with a parabola Y1 at another point n (m, n), where m ≠ N and satisfies the following conditions: M & # 178; - M + T = 0 and N & # 178; - N + T = 0 (t is a constant) ① Finding the value of K ② Let the line intersect the x-axis at point D, and p be a point in the coordinate plane. If the quadrilateral with vertices o, D, P, and M is a parallelogram, try to find the coordinates of point P (just write the coordinates of point P directly, and do not require the process of solution)