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The distance from the image vertex of quadratic function y = ax ^ 2 + K to the X axis is 3, and the shape is the same as that of parabola y = x ^ 2
The quadratic coefficient a determines the shape of the parabola, that is, the size of the opening
Because the shape is the same as the parabola y = x ^ 2, so y = x ^ 2 + K
The symmetric axis of parabola is x = 0, and y = k is obtained by substituting it into the function
So the vertex is (0, K)
The distance from the vertex to the x-axis is | K | = 3, k = ± 3
So the analytic formula is y = x ^ 2 + 3 or y = x ^ 2-3
When the parabola y = a (X-H) square + K passes through the point (- 1, - 4) and x = 1, the maximum value of Y is - 2, find the function
a(-1-h)^2+k=-4(1)
X = 1, y get the maximum value - 2
h=1 K=-2
Bring in (1)
4a-2=-4
4a=-2
a=-1/2
y=(-1/2)(x-1)^2-2
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- 1. Parabola y = - 4 (x + 3) square, when x_____ The value of function y increases with the increase of X_____ The value of function y decreases with the increase of X, and the square of parabola y = - 4 (x + 3) can be changed from the plane direction of parabola y = - 4x_____ Translation_____ Units get it!
- 2. The vertex of the parabola y = AX2 + BX + C is on the Y-axis and passes through two points - (- 1,3), (- 2,6)
- 3. Generally, we can use formula to find the vertex and symmetry axis of parabola y = ax ^ 2 + BX + C (a ≠ 0). Y = ax ^ 2 + BX + C = a [x + (B / 2a)] ^ 2+ Generally, we can find the vertex and symmetry axis of parabola y = ax ^ 2 + BX + C (a ≠ 0) y=ax^2 + bx + c =a[x+(b/2a)]^2 + (4ac-b^2)/4a, y=ax^2 + bx + c It should not be converted to (x + B / 2a) ^ 2 = B ^ 2-4ac / 4A ^ 2. How can it be converted to the above form?
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