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Assuming the vertex is 0,0 and the directrix is y=4 x^2=0
Do you have a specific vertex fraction? vertex = -b/2a wuadratic = ax^ + bx + c
when the function is in vertex form: y = a(x - h)2 + k, the point (h, k) is the vertex.
just leave the x terms to 1 side and constants to other side
The difference between standard form and vertex form is the standard form gives the coefficients(a,b,c) of the different powers of x. The vertex form gives the vertex 9hk) of the parabola as part of the equation.
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y= -5/49(x-9)^2+5
The graph of a quadratic function is always a parabola. If you put the equation (or function) into vertex form, you can read off the coordinates of the vertex, and you know the shape and orientation (up/down) of the parabola.
You complete the squares. y = ax2 + bx + c = (ax2 + b/2a)2 + c - b2/(4a2)
Assuming the vertex is 0,0 and the directrix is y=4 x^2=0
You have written it in standard form.
look for the interceptions add these and divide it by 2 (that's the x vertex) for the yvertex you just have to fill in the x(vertex) however you can also use the formula -(b/2a)
The plural form of vertex is vertices or vertexes.
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
Do you have a specific vertex fraction? vertex = -b/2a wuadratic = ax^ + bx + c
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.