Parabola: 1+12x-6x^2
Factorizing: -6(x^2 -2x -1/6)
Completing the square: -6((x-1)^2 -1 -1/6) => -6(x-1)^2 +7
Vertex of parabola is at: (1, 7)
To find the coefficient of the squared term in the parabola's equation, we can use the vertex form of a parabola, which is (y = a(x - h)^2 + k), where ((h, k)) is the vertex. Given the vertex at (3, 1), the equation starts as (y = a(x - 3)^2 + 1). Since the parabola passes through the point (4, 0), we can substitute these values into the equation: (0 = a(4 - 3)^2 + 1), resulting in (0 = a(1) + 1). Solving for (a), we find (a = -1). Thus, the coefficient of the squared term is (-1).
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To find the coefficient of the squared term in the parabola's equation, we can use the vertex form of a parabola, which is (y = a(x - h)^2 + k), where ((h, k)) is the vertex. Here, the vertex is ((-3, -1)), so the equation becomes (y = a(x + 3)^2 - 1). Given that when (y = 0), (x = 4), we can substitute these values into the equation to find (a): [0 = a(4 + 3)^2 - 1 \implies 0 = a(7^2) - 1 \implies 1 = 49a \implies a = \frac{1}{49}.] Thus, the coefficient of the squared term is (\frac{1}{49}).
You would convert it to vertex form by completing the square. You can also find the optimum value as optimum value and vertex are the same.
To find the value of a in a parabola opening up or down subtract the y-value of the parabola at the vertex from the y-value of the point on the parabola that is one unit to the right of the vertex.
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You would convert it to vertex form by completing the square. You can also find the optimum value as optimum value and vertex are the same.
The vertex of a parabola is the minimum or maximum value of the parabola. To find the maximum/minimum of a parabola complete the square: x² + 4x + 5 = x² + 4x + 4 - 4 + 5 = (x² + 4x + 4) + (-4 + 5) = (x + 2)² + 1 As (x + 2)² is greater than or equal to 0, the minimum value (vertex) occurs when this is zero, ie (x + 2)² = 0 → x + 2 = 0 → x = -2 As (x + 2)² = 0, the minimum value is 0 + 1 = 1. Thus the vertex of the parabola is at (-2, 1).
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If you want to sketch graphs you have to observe the parabola first then find the vertex afterwards you connect them and you've arrived at your answer. In order to write equations for parabolas it has to have x square in it. The standard equation for a parabola is (y - k)2 = 4a(x - h) where h and k are the x- and y-coordinates of the vertex of the parabola and 'a' is a non zero real number. This website at the related link should help, for the equation at least. A parabola is a basic U shaped graph that meets at one point called a vertex. The equation for Andy parabola must have a number being squared such as x2.
Most likely you have an equation of a parabola. The vertex of a parabola is the location where it changes from going down, to going up (a simplified explanation). Most parabolas that we think of are oriented up or down (the axis is parallel to the y axis), but they could be oriented sideways, or even at an angle. To calculate the vertex of a parabola ususally means to find the coordinates of the vertex.
To find other points on a parabola, you can use its equation, typically in the form (y = ax^2 + bx + c). By selecting different values for (x) and substituting them into the equation, you can calculate the corresponding (y) values. Alternatively, you can also use the vertex form, (y = a(x-h)^2 + k), where ((h, k)) is the vertex, to find points by choosing (x) values around the vertex. Plotting these points will help visualize the shape of the parabola.
By completing the square y = (x+3)2+1 Axis of symmetry and vertex: x = -3 and (-3, 1) Note that the parabola has no x intercepts because the discriminant is less than zero
First you need more details about the parabola. Then - if the parabola opens upward - you can assume that the lowest point of the triangle is at the vertex; write an equation for each of the lines in the equilateral triangle. These lines will slope upwards (or downwards) at an angle of 60°; you must convert that to a slope (using the tangent function). Once you have the equation of the lines and the parabola, solve them simultaneously to check at what points they cross. Finally you can use the Pythagorean Theorem to calculate the length.