your answer would actually be x=a(y-v)2+h
Y=a(x-h)^2+v
x = a(y - v)2 + h
x = a(y-v)^2 + h
would be correct for Apex, trust me I chose a different answer and got it wrong. Good luck!
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.
Is a parabola whose directrix is below its vertex.
Opening up, the vertex is a minimum.
This is the coordinate of the vertex for a parabola that opens up, defined by a positive value of x^2.
x= ay² + by + c Apex :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
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.
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.
focus , directrix
If the equation of the parabola isy = ax^2 + bx + c, then it opens above when a>0 and opens below when a<0. [If a = 0 then the equation describes a straight line, and not a parabola!].
The given terms can't be an equation without an equality sign but a negative parabola opens down wards whereas a positive parabola opens up wards.
Is a parabola whose directrix is below its vertex.
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.
The maximum.
Vertex
Opening up, the vertex is a minimum.
The maximum point.