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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.
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Find equation what parabola its vertex is 0 0 and it passes through point 2 12 express the equation in standard form?
Y=3x^2 and this is in standard form. The vertex form of a prabola is y= a(x-h)2+k The vertex is at (0,0) so we have y=a(x)^2 it goes throug (2,12) so 12=a(2^2)=4a and a=3. Now the parabola is y=3x^2. Check this: It has vertex at (0,0) and the point (2,12) is on the parabola since 12=3x2^2
Find the vertex and equation of the directri for y2 equals -32x?
y2 = 32x y = ±√32x the vertex is (0, 0) and the axis of symmetry is x-axis or y = 0
How do you graph y equals x squared minus 4?
its a simple parobola symmetric about y axis, having its vertex at (0,-4). we can make its graph by changing its equation in standard form so that we can get its different standard points like vertex, focus, etc.
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?
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
Find the equation of the axis symmetry and the coordinates of the vertex of the graph of each function for y equals 2x plus 4?
I'm assuming that you meant y = 2(x^2) +4. If it were only y = 2x +4, then this would be a linear equation and not a parabola. Anyways, use the equation x = -b/2a to find the x-value of your vertex AND your axis of symmetry. (Given the standard equation y = a(x^2) + bx + c) So, x = -0/2(2) - x = 0 (Axis of Symmetry) Now plug 0 back into your equation to find your y-value of your vertex. y = 2(0^2)+4 y=0 + 4 y = 4 Therefore Vertex = (0,4)
What is the equation of a parabola with a vertex at 0 0 and a focus at 0 6?
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
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.
Find equation what parabola its vertex is 0 0 and it passes through point 2 12 express the equation in standard form?
Y=3x^2 and this is in standard form. The vertex form of a prabola is y= a(x-h)2+k The vertex is at (0,0) so we have y=a(x)^2 it goes throug (2,12) so 12=a(2^2)=4a and a=3. Now the parabola is y=3x^2. Check this: It has vertex at (0,0) and the point (2,12) is on the parabola since 12=3x2^2
How do you solve parabola equation having lb focus at 0 3?
There is not enough information. You need either the directrix or vertex (or some other item of information).
Find the vertex and equation of the directri for y2 equals -32x?
y2 = 32x y = ±√32x the vertex is (0, 0) and the axis of symmetry is x-axis or y = 0
How do you graph y equals x squared minus 4?
its a simple parobola symmetric about y axis, having its vertex at (0,-4). we can make its graph by changing its equation in standard form so that we can get its different standard points like vertex, focus, etc.
What is the standard form of the equation of the parabola with vertex 00 and directrix y4?
Assuming the vertex is 0,0 and the directrix is y=4 x^2=0
What is the coefficient of the squared term in the parabola's equation when the vertex is at 2 -1 and the point 5 0 is on it?
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
What does 4 stands for in equation of parabola square of square of y equals 4ax?
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 What is the coefficient of the squared term in the parabolas equation?
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
Find the equation of the axis symmetry and the coordinates of the vertex of the graph of each function for y equals 2x plus 4?
I'm assuming that you meant y = 2(x^2) +4. If it were only y = 2x +4, then this would be a linear equation and not a parabola. Anyways, use the equation x = -b/2a to find the x-value of your vertex AND your axis of symmetry. (Given the standard equation y = a(x^2) + bx + c) So, x = -0/2(2) - x = 0 (Axis of Symmetry) Now plug 0 back into your equation to find your y-value of your vertex. y = 2(0^2)+4 y=0 + 4 y = 4 Therefore Vertex = (0,4)
What is the vertex of fx equals 5xsquared?
The vertex is at the origin of coordinates ... the point (0, 0).
