On quadratic equation of one variable from the point of view of function
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Mathematics elementary three function one variable quadratic equation question
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Given the parabola y = (K + 1) x ^ 2 + 2kx + K-2, ask why K is the value of K. the parabola and X intersect at two points and intersect at one point without intersection
Why is there two answers to zero? Can you tell me your answer.
Parabola y = (K + 1) x ^ 2 + 2kx + K-2, a = K + 1, B = 2K, C = K-2,
Discriminant = △ = B ^ 2-4ac
=(2k)^2-4(k+1)(k-2)
=4k^2-4(k^2-k-2)
=4k^2-4k^2+4k+8
=4k+8,
When △ > 0, i.e. 4K + 8 > 0, k > - 2, it intersects the X axis at two points
When △ = 0, that is, 4K + 8 = 0, k = - 2, it intersects with the X axis at a point
When △
Quadratic function and quadratic equation of one variable
As shown in the figure, the parabola y = ax ^ 2 + BX + C intersects the x-axis at points a and B. it intersects the y-axis at point C. If ob = OC = 1 / 2oa, the value of B is obtained
The vertex of the parabola is open in the second quadrant, a is on the left of Y, B is on the right of Y
Don't copy other people's
X = 0, y = 0 + 0 + C = C, so C (0, c), obviously C is above the origin, so c > 0, then ob = OC = CB (C, 0) OA = 2ob, and a is to the left of Y, so a (- 2C, 0) so C and - 2C are two of the equations ax ^ 2 + BX + C = 0, followed by c * (- 2C) = C / A, so - 2C ^ 2 = C / A, a = - 1 / (2C) C + (- 2C) = - B / A, - C = - B / [- 1 / (2C)] = 2BC, so B = - 1 /
Write out the expression of a linear function in the first, second and third quadrant of an image______ .
If ∵ passes through one or three quadrants, the coefficient of ∵ x is greater than 0, that is, k > 0; if ∵ passes through the second quadrant, it means that the constant term is greater than 0, b > 0
If y is inversely proportional to x, a point P (a, b) in the fourth quadrant is on the image of this function, and a, B are the solutions of the equation (x-1) square = 4
First solve equation 3 or - 1
A point P (a, b) in the fourth quadrant is on the image of this function, and a, B are the solution of equation (x-1) square = 4
Of
a=3 b=-1
Let y = K / X (k is not equal to 0)
So y = - 3 / X
Write out a function expression that the image passes through the point (- 1, - 1) and does not pass through the first quadrant______ .
It can be a linear function y = KX + B, or a quadratic function y = AX2 + BX + C. ∵ crossing point (- 1, - 1) ∵ the answer is not unique, such as y = - X-2 or y = - X2, so fill in the blanks: y = - X-2 or Y = - x2
Which function image is increasing in the first quadrant, and the maximum value of the function is not more than 1, that is, y = 1 as the asymptote
You'd better provide the function image,
Of course, it's not necessary that the asymptote is y = 1, or it can be slightly smaller, such as y = 0.8 and so on, not too small.
Also, I have a point to add. It's better not to translate a function. I hope it's such a function. It can be transformed by some function, but not by translation.
The increasing range is the independent variable all the way to positive infinity.
Function y = - 1 / x + 1
Image processing of quadratic function y = 2x ^ 2-4x__ quadrant
y=2x^2-4x=2x(x-2)=2(x-1)^2-2,
So the graph of this function passes through the points (0,0), (2,0) and the coordinates of the lowest point are (1, - 2), so the graph is easy to draw. Then we can see that the graph passes through quadrants 1, 2 and 4
The vertex of image with quadratic function y = 3 (x + 1) ^ 2-5____ Quadrant
Vertex (- 1, - 5)
The third__ 3_ quadrant
In quadratic function, if a is greater than 0, B is greater than 0, C is equal to 0, the image does not pass through the quadrant
In the quadratic function y = ax square + BX + C, if a is greater than 0b and greater than 0C is equal to 0, the image does not pass through the quadrant
It doesn't pass through the fourth quadrant because the vertex of y = ax ^ 2 is always on y = BX and the opening is upward