Y=3x^2 and this is in standard form. The vertex form of a prabola is y= a(x-h)2+k The vertex is at (0,0) so we have y=a(x)^2 it goes throug (2,12) so 12=a(2^2)=4a and a=3. Now the parabola is y=3x^2. Check this: It has vertex at (0,0) and the point (2,12) is on the parabola since 12=3x2^2
y = 8/49*x2
please help
A parabola that opens upward is a U-shaped curve where the vertex is the lowest point on the graph. It can be represented by the general equation y = ax^2 + bx + c, where a is a positive number. The axis of symmetry is a vertical line passing through the vertex, and the parabola is symmetric with respect to this line. The focus of the parabola lies on the axis of symmetry and is equidistant from the vertex and the directrix, which is a horizontal line parallel to the x-axis.
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"From the geometric point of view, the given point is the focus of the parabola and the given line is its directrix. It can be shown that the line of symmetry of the parabola is the line perpendicular to the directrix through the focus. The vertex of the parabola is the point of the parabola that is closest to both the focus and directrix."-http://www.personal.kent.edu/~rmuhamma/Algorithms/MyAlgorithms/parabola.htm"A line perpendicular to the axis of symmetry used in the definition of a parabola. A parabola is defined as follows: For a given point, called the focus, and a given line not through the focus, called the directrix, a parabola is the locus, or set of points, such that the distance to the focus equals the distance to the directrix."-http://www.mathwords.com/d/directrix_parabola.htm
The primary focal chord of a parabola is a line segment that passes through the focus of the parabola and has its endpoints on the parabola itself. For a standard parabola defined by the equation (y^2 = 4px), the focus is located at the point ((p, 0)). The primary focal chord is unique in that it is perpendicular to the axis of symmetry of the parabola and is the longest chord that can be drawn through the focus.
y = 8/49*x2
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To find the equation of a parabola with vertex at ((-3, 0)) that passes through the point ((3, 18)), we can use the vertex form of a parabola, (y = a(x + 3)^2). To determine the value of (a), substitute the point ((3, 18)) into the equation: [ 18 = a(3 + 3)^2 \implies 18 = a(6)^2 \implies 18 = 36a \implies a = \frac{1}{2}. ] Thus, the equation of the parabola is (y = \frac{1}{2}(x + 3)^2).
Assuming that a is the leading coefficient of the equation of the parabola, changing it from positive to negative will reflect the parabola along a horizontal line through its minimum - which will then become its maximum.
To convert a vertex form equation of a parabola, given as ( y = a(x - h)^2 + k ), to standard form ( y = ax^2 + bx + c ), expand the squared term: ( (x - h)^2 = x^2 - 2hx + h^2 ). Then, multiply through by ( a ) and combine like terms: ( y = ax^2 - 2ahx + (ah^2 + k) ). The coefficients ( a ), ( b = -2ah ), and ( c = ah^2 + k ) represent the standard form parameters.
please help
the equation of a parabola is: y = a(x-h)^2 + k *h and k are the x and y intercepts of the vertex respectively * x and y are the coordinates of a known point the curve passes though * solve for a, then plug that a value back into the equation of the parabola with out the coordinates of the known point so the equation of the curve with the vertex at (0,3) passing through the point (9,0) would be.. 0 = a (9-0)^2 + 3 = 0 = a (81) + 3 = -3/81 = a so the equation for the curve would be y = -(3/81)x^2 + 3
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).
Suppose the equation of the parabola is y = ax2 + bx + c where a, b, and c are constants, and a ≠0. The roots of the parabola are given by x = [-b ± sqrt(D)]/2a where D is the discriminant. Rather than solve explicitly for the coordinates of the vertex, note that the vertical line through the vertex is an axis of symmetry for the parabola. The two roots are symmetrical about x = -b/2a so, whatever the value of D and whether or not the parabola has real roots, the x coordinate of the vertex is -b/2a. It is simplest to substitute this value for x in the equation of the parabola to find the y-coordinate of the vertex, which is c - b2/2a.
The latus rectum of a parabola is a segment with endpoints on the parabola passing through the focus and parallel to the directrix.
A parabola that opens upward is a U-shaped curve where the vertex is the lowest point on the graph. It can be represented by the general equation y = ax^2 + bx + c, where a is a positive number. The axis of symmetry is a vertical line passing through the vertex, and the parabola is symmetric with respect to this line. The focus of the parabola lies on the axis of symmetry and is equidistant from the vertex and the directrix, which is a horizontal line parallel to the x-axis.