(3,5)
i think that the range and the domain of a parabola is the coordinates of the vertex
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.
-3
Without an equality sign it can not be considered to be an equation and the - or + value of 9x has not been given
A parabola is a type of graph that is not linear, and mostly curved. A parabola has the "x squared" sign in it's equation. A parabola is not only curved, but all the symmetrical. The symmetrical point, the middle of the parabola is called the vertex. You can graph this graph with the vertex, x-intercepts and a y-intercept. A parabola that has a positive x squared would be a smile parabola, and the one with the negative x squared would be a frown parabola. Also, there are the parabolas that are not up or down, but sideways Those parabolas have x=y squared, instead of y = x squared.
We will be able to identify the answer if we have the equation. We can only check on the coordinates from the given vertex.
i think that the range and the domain of a parabola is the coordinates of the vertex
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.
0,0
Most likely you have an equation of a parabola. The vertex of a parabola is the location where it changes from going down, to going up (a simplified explanation). Most parabolas that we think of are oriented up or down (the axis is parallel to the y axis), but they could be oriented sideways, or even at an angle. To calculate the vertex of a parabola ususally means to find the coordinates of the vertex.
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.
-2
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
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
3
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.
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.