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How do you write an equation of a parabola with vertex at the origin and the given focus 60?

To write the equation of a parabola with its vertex at the origin (0, 0) and a focus at (0, 60), you first identify the orientation of the parabola. Since the focus is above the vertex, the parabola opens upwards. The standard form of the equation for a parabola that opens upwards is ( y = \frac{1}{4p}x^2 ), where ( p ) is the distance from the vertex to the focus. Here, ( p = 60 ), so the equation becomes ( y = \frac{1}{240}x^2 ).


What is a quadratic equation in vertex form for a parabola with vertex (11 -6)?

A quadratic equation in vertex form is expressed as ( y = a(x - h)^2 + k ), where ((h, k)) is the vertex of the parabola. For a parabola with vertex at ((11, -6)), the equation becomes ( y = a(x - 11)^2 - 6 ). The value of (a) determines the direction and width of the parabola. Without additional information about the parabola's shape, (a) can be any non-zero constant.


What is the equation for the parabola with the vertex -3.0 that passes through the point 318?

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).


The vertex of this parabola is at 3 1 When the y-value is 0 the x-value is 4 What is the coefficient of the squared term in the parabolas equation?

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).


A parabola has a vertex at -3 -2 what is its equation?

-1

Related Questions

The vertex of the parabola below is at the point 4 -1 which equation be this parabola's equation?

5


How do you write an equation of a parabola with vertex at the origin and the given focus 60?

To write the equation of a parabola with its vertex at the origin (0, 0) and a focus at (0, 60), you first identify the orientation of the parabola. Since the focus is above the vertex, the parabola opens upwards. The standard form of the equation for a parabola that opens upwards is ( y = \frac{1}{4p}x^2 ), where ( p ) is the distance from the vertex to the focus. Here, ( p = 60 ), so the equation becomes ( y = \frac{1}{240}x^2 ).


The vertex of the parabola below is at the point (-4-2) which equation below could be one for parabola?

-2


What is a quadratic equation in vertex form for a parabola with vertex (11 -6)?

A quadratic equation in vertex form is expressed as ( y = a(x - h)^2 + k ), where ((h, k)) is the vertex of the parabola. For a parabola with vertex at ((11, -6)), the equation becomes ( y = a(x - 11)^2 - 6 ). The value of (a) determines the direction and width of the parabola. Without additional information about the parabola's shape, (a) can be any non-zero constant.


What is the equation for the parabola with the vertex -3.0 that passes through the point 318?

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).


The vertex of this parabola is at 3 1 When the y-value is 0 the x-value is 4 What is the coefficient of the squared term in the parabolas equation?

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).


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)


A parabola has a vertex at -3 -2 what is its equation?

-1


The vertex of this parabola is at -2 -3 When the y-value is -2 the x-value is -5 What is the coefficient of the squared term in the parabola's equation?

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.


What is the equation of the parabola with focuse (07) and the directrix y1?

To find the equation of the parabola with focus at (0, 7) and directrix ( y = 1 ), we first determine the vertex, which is the midpoint between the focus and the directrix. The vertex is at ( (0, 4) ). The distance from the vertex to the focus is 3, so the parabola opens upward. The equation of the parabola can be expressed as ( (x - h)^2 = 4p(y - k) ), where ( (h, k) ) is the vertex and ( p ) is the distance from the vertex to the focus. Thus, the equation is ( x^2 = 12(y - 4) ).


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.


The vertex of the parabola below is at the point -3 -5 Which of the equations below could be the equation of this parabola?

2