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
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
Recall that the graph of a linear equation in two variables is a line. The equation y = ax^2 + bx + c, where a, b, and c are real numbers and a is different than 0 represents a quadratic function. Its graph is a parabola, a smooth and symmetric U-shape. 1. The axis of symmetry is the line that divides the parabola into two matching parts. Its equation is x = -b/2a 2. The highest or lowest point on a parabola is called the vertex (also called a turning point). Its x-coordinate is the value of -b/2a. If a > 0, the parabola opens upward, and the vertex is the lowest point on the parabola. The y-coordinate of the vertex is the minimum value of the function. If a < 0, the parabola opens downward, and the vertex is the highest point on the parabola. The y-coordinate of the vertex is the maximum value of the function. 3. The x-intercepts of the graph of y = ax^2 + bx + c are the real solutions to ax^2 + bx + c = 0. The nature of the roots of a quadratic function can be determined by looking at its graph. If you see that there are two x-intercepts on the graph of the equation, then the equation has two real roots. If you see that there is one x-intercept on the graph of the equation, then the equation has one real roots. If you see that the graph of the equation never crosses the x-axis, then the equation has no real roots. The roots can be used further to determine the factors of the equation, as (x - r1)(x -r2) = 0
Vertex = (3, - 2)Put in vertex form.(X - 3)2 + 2X2 - 6X + 9 + 2 = 0X2 - 6X + 11 = 0=============The coefficeint of the squared term is 1. My TI-84 confirms the (4, 3) intercept of the parabola and the 11 Y intercept shown by the function.
Did you mean a parabola with equation y=3x^2? The line of symmetry is x=0 or the y-axis.
10
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.
The equation of a parabola that opens left or right with its vertex at the point ((h, v)) is given by ((y - v)^2 = 4p(x - h)), where (p) is the distance from the vertex to the focus. If (p > 0), the parabola opens to the right, and if (p < 0), it opens to the left.
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) ).
x2 = 16y The standard formula for a parabola with its vertex at the origin (0, 0) and a given focus (and the y-axis as an axis of symmetry) is as follows: x2 = 4cy In this case, the c is the y value of the focus. The focus in this case was (0, 4), and the y value in the focus is 4. That makes the c = 4. Further, that makes the equation for this parabola x2 = 4 (c)y = 4 (4)y = 16y Given that the vertex was the origin, (0, 0), and the focus is (0, 4), we can conclude that the axis of symmetry is the y-axis because the y value of the focus is 0. We can also conclude that the parabola opens up, because the focus has a positive y value.
There is not enough information. You need either the directrix or vertex (or some other item of information).
A parabola with an equation, y2 = 4ax has its vertex at the origin and opens to the right. It's not just the '4' that is important, it's '4a' that matters. This type of parabola has a directrix at x = -a, and a focus at (a, 0). By writing the equation as it is, the position of the directrix and focus are readily identifiable. For example, y2 = 2.4x doesn't say a great deal. Re-writing the equation of the parabola as y2 = 4*(0.6)x tells us immediately that the directrix is at x = -0.6 and the focus is at (0.6, 0)
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
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 ).
The equation that describes a parabola opening left or right with its vertex at the point ((h, k)) is given by ((y - k)^2 = 4p(x - h)), where (p) determines the direction and width of the parabola. If (p > 0), the parabola opens to the right, while if (p < 0), it opens to the left. Here, ((h, k)) represents the vertex coordinates.
Assuming the vertex is 0,0 and the directrix is y=4 x^2=0
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
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. Here, the vertex is ((-3, -1)), so the equation becomes (y = a(x + 3)^2 - 1). Given that when (y = 0), (x = 4), we can substitute these values into the equation to find (a): [0 = a(4 + 3)^2 - 1 \implies 0 = a(7^2) - 1 \implies 1 = 49a \implies a = \frac{1}{49}.] Thus, the coefficient of the squared term is (\frac{1}{49}).