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What is the equation of a parabola with vertex at 1 -3 and focus at 2 -3?
Wiki User
∙ 8y ago
For a parabola with an axis of symmetry parallel to the x-axis, the equation of a parabola is given by:
(y - k)² = 4p(x - h)
Where the vertex is at (h, k), and the distance between the focus and the vertex is p (which can be calculated as p = x_focus - x_vertex).
For the parabola with vertex (1, -3) and focus (2, -3) this gives:
h = 1
k = -3
p = 2 - 1 = 1
→ parabola is:
(y - -3)² = 4×1(x - 1)
→ (y + 3)² = 4(x - 1)
This can be expanded to:
4x = y² + 6y + 13
or
x = (1/4)y² + (3/2)y + (13/4)
Wiki User
∙ 8y ago
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A parabola You can find the peak of this parabola by taking it's derivative and finding the point where it's slope is equal to zero: y = 3x2 + x - 2 y' = 6x + 1 Let y' = 0 0 = 6x + 1 x = -1/6 Now find the y co-ordinate: y = 3(-1/6)2 + (-1/6) - 2 y = 3/18 - 1/6 - 2 y = -2 So this parabola's vertex will be located at the point (-1/6, -2).
What is the standard equation for vertex at origin opens down 1 and 76 units between the vertex and focus?
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5
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3
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-1
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The vertex of the parabola below is at the point -2 1 Which of the equations below could be this parabolas equation?
Go study
What is the coefficient of the squared term in the parabola's equation when the vertex is at -2 -3 and the point -1 -5 is on it?
A parabola with vertex (h, k) has equation of the form: y = a(x - h)² + k → vertex (k, h) = (-2, -3), and a point on it is (-1, -5) → -5 = a(-1 - -2)² + -3 → -5 = a(1)² - 3 → -5 = a - 3 → a = -2 → The coefficient of the x² term is -2.
What is the coefficient of the squared term in the parabola's equation when the vertex is at 3 5 and the point -1 6 is on it?
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Vertex of a parabola?
The vertex of a parabola is found by using the solution of the equation -b/2a and putting it into the quadratic equation. a is the coefficient of x^2. b is the coefficient of the other x in the equation. Ex. y=2x^2+2x+1 -b/2a = -2/2(2) = -1/2 Now put -1/2 in the place of every x in the equation. y=2(-1/2)^2+2(-1/2)+1 The vertex is (-1/2, 1/2)
