Use this form: y= a(x-h)² + k ; plug in the x and y coordinates of the vertex into (h,k) and then the other point coordinates into (x,y) and solve for a.
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The standard equation for a Parabola with is vertex at the origin (0,0) is, x2 = 4cy if the parabola opens vertically upwards/downwards, or y2 = 4cx when the parabola opens sideways. As the focus is at (0,6) then the focus is vertically above the vertex and we have an upward opening parabola. Note that c is the distance from the vertex to the focus and in this case has a value of 6 (a positive number). The equation is thus, x2 = 4*6y = 24y
The vertex of this parabola is at -2 -3 When the y-value is -2 the x-value is -5. The coefficient of the squared term in the parabola's equation is -3.
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please help
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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
The standard equation for a Parabola with is vertex at the origin (0,0) is, x2 = 4cy if the parabola opens vertically upwards/downwards, or y2 = 4cx when the parabola opens sideways. As the focus is at (0,6) then the focus is vertically above the vertex and we have an upward opening parabola. Note that c is the distance from the vertex to the focus and in this case has a value of 6 (a positive number). The equation is thus, x2 = 4*6y = 24y
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Finding the vertex of the parabola is important because it tells you where the bottom (or the top, for a parabola that 'opens' downward), and thus where you can begin graphing.
Vertex form is denoted by: y=a(x-h)2+k Where (h,k) is the vertex. So, we have: y=a(x-2)2+3 (This super\subscript thing is annoying). Plug in the values for x and y for the point in the equation and you have your answer.
We will be able to identify the answer if we have the equation. We can only check on the coordinates from the given vertex.
The vertex of this parabola is at -2 -3 When the y-value is -2 the x-value is -5. The coefficient of the squared term in the parabola's equation is -3.
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
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