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Assuming the vertex is 0,0 and the directrix is y=4 x^2=0
when the function is in vertex form: y = a(x - h)2 + k, the point (h, k) is the vertex.
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).
So you need something like this: y = a*(x - 4)² + 3. This will make the vertex be at (4,3). Then it looks like you have another point on the parabola (3,5). Plug that in and solve for a. 5 = a*(3-4)² + 3. This becomes 5 = a + 3, so a=2, then the equation is: y = 2*(x - 4)² + 3
By inspection you should be able to see that this is a parabola with a vertex of this. (0, 0) There is no form for this function as there is no linear term.
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.
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.
Assuming the vertex is 0,0 and the directrix is y=4 x^2=0
In the equation y x-5 2 plus 16 the standard form of the equation is 13. You find the answer to this by finding the value of X.
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
This is called the 'standard form' for the equation of a parabola:y =a (x-h)2+vDepending on whether the constant a is positive or negative, the parabola will open up or down.
please help
the standard form of the equation of a parabola is x=y2+10y+22
There are two standard form of parabola: y2 = 4ax & x2 = 4ay, where a is a real number.
Normally a quadratic equation will graph out into a parabola. The standard form is f(x)=a(x-h)2+k
The focus of a parabola is a fixed point that lies on the axis of the parabola "p" units from the vertex. It can be found by the parabola equations in standard form: (x-h)^2=4p(y-k) or (y-k)^2=4p(x-h) depending on the shape of the parabola. The vertex is defined by (h,k). Solve for p and count that many units from the vertex in the direction away from the directrix. (your focus should be inside the curve of your parabola)
There are two forms in which a quadratic equation can be written: general form, which is ax2 + bx + c, and standard form, which is a(x - q)2 + p. In standard form, the vertex is (q, p). So to find the vertex, simply convert general form into standard form.The formula often used to convert between these two forms is:ax2 + bx + c = a(x + b/2a)2 + c - b2/4aSubstitute the variables:-2x2 + 12x - 13 = -2(x + 12/-4)2 -13 + 122/-8-2x2 + 12x - 13 = -2(x - 3)2 + 5Since the co-ordinates of the vertex are equal to (q, p), the vertex of the parabola defined by the equation y = -2x2 + 12x - 13 is located at point (3, 5)