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How do you find an equation of a parabola using the vertex and a point it passes through?
DresXsage
I would use the alternate parabolic equation (y-k)2 = a (x-h). You can plug in the coordinates of the vertex (h, k)--h is the x coordinate of the vertex and k is the y coordinate of the vertex.
Now you are left with an equation with an x and y, which are fine, but also an a, which we still have to get rid of. The a describes how steeply the parabola increases. To find the actual number, plug the other point you were given into the equation (where the x and y are). How you are left with one equation and one variable, so you can solve for a = some number.
Now return to the beginning of the previous step, when you had and x, y, and a in your equation. Keep the x and y in their original positions, but replace a with the number you just found.
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What is the equation of a parabola with the vertex of 2 -1?
3
What is the parabola equation?
the equation of a parabola is: y = a(x-h)^2 + k *h and k are the x and y intercepts of the vertex respectively * x and y are the coordinates of a known point the curve passes though * solve for a, then plug that a value back into the equation of the parabola with out the coordinates of the known point so the equation of the curve with the vertex at (0,3) passing through the point (9,0) would be.. 0 = a (9-0)^2 + 3 = 0 = a (81) + 3 = -3/81 = a so the equation for the curve would be y = -(3/81)x^2 + 3
What is the coefficient of the x² term of the parabola with a vertex at -3 -1 and passes through the point 4 0?
A parabola with vertex (h, k) has equation: y = a(x - h)² + k With vertex (-3, -1) this becomes: y = a(x - -3)² + -1 = a(x + 3)² - 1 The point (4, 0) is on this parabola so: 0 = a(4 + 3)² - 1 → 7²a = 1 → a = 1/49 Thus the coefficient of x² is 1/49.
What could be the equation of the parabola centered at the vertex?
f(x)=x^2
How are the vertices of the parabolas related to the equation of the quadratic function?
Suppose the equation of the parabola is y = ax2 + bx + c where a, b, and c are constants, and a ≠0. The roots of the parabola are given by x = [-b ± sqrt(D)]/2a where D is the discriminant. Rather than solve explicitly for the coordinates of the vertex, note that the vertical line through the vertex is an axis of symmetry for the parabola. The two roots are symmetrical about x = -b/2a so, whatever the value of D and whether or not the parabola has real roots, the x coordinate of the vertex is -b/2a. It is simplest to substitute this value for x in the equation of the parabola to find the y-coordinate of the vertex, which is c - b2/2a.
What is an equation of the parabola in vertex form that passes through (13 8) and has vertex (3 2).?
please help
What is the equation of vertex in parabola?
0,0
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 find the equation of a parabola if you know the vertex and a point it passes through?
Use this form: y= a(x-h)² + k ; plug in the x and y coordinates of the vertex into (h,k) and then the other point coordinates into (x,y) and solve for a.
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 an equation of the parabola in vertex form that passes through (13 8)(13 8) and has vertex (3 2)(3 2).?
Vertex form is denoted by: y=a(x-h)2+k Where (h,k) is the vertex. So, we have: y=a(x-2)2+3 (This super\subscript thing is annoying). Plug in the values for x and y for the point in the equation and you have your answer.
What is the equation of a parabola with the vertex of 2 -1?
3
How does finding the vertex of a parabola help you when graphing a quadratic equation?
Finding the vertex of the parabola is important because it tells you where the bottom (or the top, for a parabola that 'opens' downward), and thus where you can begin graphing.
What is the parabola equation?
the equation of a parabola is: y = a(x-h)^2 + k *h and k are the x and y intercepts of the vertex respectively * x and y are the coordinates of a known point the curve passes though * solve for a, then plug that a value back into the equation of the parabola with out the coordinates of the known point so the equation of the curve with the vertex at (0,3) passing through the point (9,0) would be.. 0 = a (9-0)^2 + 3 = 0 = a (81) + 3 = -3/81 = a so the equation for the curve would be y = -(3/81)x^2 + 3
What is the y-coordinate of the vertex of the parabola that is given by the equation below?
We will be able to identify the answer if we have the equation. We can only check on the coordinates from the given vertex.
The vertex of the parabola below is at the point (-4-2) which equation below could be one for parabola?
-2
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.
