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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)

Q: What is the equation of a parabola with vertex at 1 -3 and focus at 2 -3?

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It is 1/16.

The minimum value of the parabola is at the point (-1/3, -4/3)

The axis of symmetry is x = -2.

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.

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).

Related questions

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.

5

3

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.

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).

-1

The vertex of this parabola is at -3 -1 When the y-value is 0 the x-value is 4. The coefficient of the squared term in the parabolas equation is 7

A parabola with vertex (h, k) has equation of the form: y = a(x - h)² + k → vertex (k, h) = (2, -1), and a point on it is (5, 0) → 0 = a(5 - 2)² + -1 → 0 = a(3)² -1 → 1 = 9a → a = 1/9 → The coefficient of the x² term is 1/9

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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.

A parabola with vertex (h, k) has equation of the form: y = a(x - h)² + k → vertex (k, h) = (3, 5), and a point on it is (-1, 6) → 6 = a(-1 - 3)² + 5 → 6 = a(-4)² + 5 → 1 = 16a → a = 1/16 → The coefficient of the x² term is 1/16

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)