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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)
What else can I help you with?
When vertex of this parabola is at (35) . When the y-value is 6 the x-value is -1. what is the coefficient of the squared term in the parabolas equation?
It is 1/16.
What is the vertex of the parabola when y equals 3x squared plus 2x minus 1?
The minimum value of the parabola is at the point (-1/3, -4/3)
What is the axis of symmetry for the parabola with vertex (-2 -4) and directrix y 1?
The axis of symmetry is x = -2.
How do you graph a quadratic equation?
A quadratic equation is an equation with the form: y=Ax2+Bx+C The most important point when graphing a parabola (the shape formed by a quadratic) is the vertex. The vertex is the maximum or minimum of the parabola. The x value of the vertex is equal to -B/(2A). Once you have the x value, just plug it back into the original equation to get the corresponding y value. The resulting ordered pair is the location of the vertex. A parabola will be concave up (pointed downward) if A is +. It will be concave down (pointed upward) if A is -. It is often helpful to find the zeroes of a function when graphing. This can be done by factoring or using the quadratic formula. For every n units away from the vertex on the x-axis, the corresponding y value goes up (or down) by n2*A. Parabolas are symetrical along the vertex, which means that if one point is n units from the vertex, the point -n units from the vertex has the same y value. As an example take the following quadratic: 2x2-8x+3 A=2, B=-8, and C=3 The x value of the vertex is -B/2A=-(-8)/(2*2)=2 By plugging 2 into the original equation we get that the vertex is at (2,-5) 3 units to the right (x=5) has a y value of -5+32*2=13. This means that 3 units to the left (x=-1) has the same y value (-1,13). If you need a clearer explanation, ask a math teacher.
What is the shape of the curve y equals 3x squared plus x minus 2?
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).
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 the standard equation for vertex at origin opens down 1 and 76 units between the vertex and focus?
Since the vertex is at the origin and the parabola opens downward, the equation of the parabola is x2 = 4py, where p < 0, and the axis of symmetry is the y-axis. So the focus is at y-axis at (0, p) and the directrix equation is y = -p. Now, what do you mean with 1 and 76 units? 1.76 units? If the distance of the vertex and the focus is 1.76 units, then p = -1.76, thus 4p = -7.04, then the equation of the parabola is x2 = -7.04y.
What is the focus of the parabola y equals 4x2?
The equation ( y = 4x^2 ) represents a parabola that opens upwards. To find the focus, we can rewrite it in the standard form ( y = 4p(x - h)^2 + k ), where ( (h, k) ) is the vertex and ( p ) is the distance from the vertex to the focus. Here, the vertex is at the origin ( (0, 0) ) and ( 4p = 4 ), so ( p = 1 ). Thus, the focus of the parabola is located at the point ( (0, 1) ).
The vertex of the parabola below is at the point 4 -1 which equation be this parabola's equation?
5
What is the equation of a parabola with the vertex of 2 -1?
3
What is the focus of the parabola y 4x2?
The equation of the parabola ( y = 4x^2 ) can be rewritten in the standard form ( y = 4p(x - h)^2 + k ), where ( (h, k) ) is the vertex. Here, it is clear that the vertex is at the origin (0, 0) and ( 4p = 4 ), giving ( p = 1 ). The focus of the parabola is located at ( (h, k + p) ), so the focus is at the point ( (0, 1) ).
The vertex of this parabola is at 5 5 When the x-value is 6 the y-value is -1 What is the coefficient of the squared expression in the parabola's equation?
The vertex of this parabola is at 5 5 When the x-value is 6 the y-value is -1. The coefficient of the squared expression in the parabola's equation is -6.
How do you get directrix of a parabola?
The directrix of a parabola can be found using its standard form equation. For a parabola that opens upwards or downwards, given by (y = ax^2 + bx + c), the directrix is located at (y = k - \frac{1}{4p}), where (k) is the vertex's y-coordinate and (p) is the distance from the vertex to the focus. For a parabola that opens sideways, the directrix is given by (x = h - \frac{1}{4p}), where (h) is the vertex's x-coordinate. The value of (p) can be determined based on the coefficients of the quadratic equation.
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).
The co-ordinates of the focus of Parabola x minus 3 power 2 is equal to 4 y minus 2 is Answer is 3 3 how?
The standard equation for an upward opening parabola with its vertex at (f,g) is (x - f)2 = 4c(y - g), where c is the focal length, that is the distance of the focus from the vertex. Putting the equation shown in this form we have :- (x - 3)2 = 4 * 1(y - 2) . . . thus c = 1 The vertex is at (3,2) and as the focal length is 1 (and this is positive) then the coordinates of the focus are (3, [2 + 1]) = (3,3).
What could be the equation of a parabola with its vertex at (-36).?
The equation of a parabola with its vertex at the point (-36, k) can be expressed in the vertex form as ( y = a(x + 36)^2 + k ), where ( a ) determines the direction and width of the parabola. If the vertex is at (-36), the x-coordinate is fixed, but the y-coordinate ( k ) can vary depending on the specific position of the vertex. If you'd like a specific example, assuming ( k = 0 ) and ( a = 1 ), the equation would be ( y = (x + 36)^2 ).
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).
