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Unfortunately, it is not possible to know what the question was, so here are some possible answers 4 x 6y = 24y These questions have a range of solutions which could be plotted on a graph. It is difficult to show the graphic representation here. See the related links to plot the answers. 4X x 6Y --> 6Y = 1/4X --> Y = (1/4X)/6 4X x 6 + Y --> Y = - (4X x 6)
one vertex: 3 two vertices: 6 three vertices: 8 total 17
A cube has no vertex
Vertex of a triangle is any of its 3 corners and the plural of vertex is vertices
As the second line is parallel to the first then the slopes of both lines are the same. 4x + y - 1 = 0 : y = -4x + 1 : Therefore the slope is -4. Using the standard slope-intercept equation of y = mx + c for the second line gives :- 2 = -4x1 + c : 2 = -4 + c : c = 2 + 4 = 6. Therefore the equation of the second line is y = -4x + 6. (Or y + 4x - 6 = 0)
y = 2x2 + 4x - 3 This equation describes a parabola. Because it's first term is positive, we know that it goes infinitely upward, and has a minimum that occurs at it's vertex. You can find it's vertex by taking it's derivative and solving for zero: y' = 4x + 4 0 = 4x + 4 0 = x + 1 x = -1 y = (-1)2 + 4(-1) - 3 y = 1 - 4 - 3 y = -6 So the vertex is at (-1, -6), which means that y ≥ -6
Example function.Y = X2 - 4X + 5set to 0X2 - 4X + 5 = 0X2 - 4X = - 5Now, halve the linear coefficient, square it and add it to both sidesX2 - 4X + 4 = - 5 + 4gather terms on the right and factor on the left(X - 2)2 = -1==============Vertex form.(2, - 1)=======Vertex.
2x2= 4
Y = X2 - 4X - 5set to zeroX2 - 4X - 5 = 0X2 - 4X = 5halve the linear term ( - 4 ) then square it and add that result to both sidesX2 - 4X + 4 = 5 + 4factor on the left and gather terms together on the right(X - 2)2 = 9(X - 2)2 - 9 = 0==============vertex form(2, - 9)======vertex
y=4x-12-3 is the equation of a straight line. It does not have a vertex. Did you mean y=x squared - 12x - 3 ?
Assuming the missing symbol there is an equals sign, then we have: y - 2x2 - 4x = 4 We can find it's vertex very easily by solving for y, and finding where it's derivative equals zero: y = 2x2 + 4x + 4 y' = 4x + 4 0 = 4x + 4 x = -1 So the vertex occurs Where x = -1. Now we can plug that back into the original equation to find y: y = 2x2 + 4x + 4 y = 2 - 4 + 4 y = 2 So the vertex is at the point (-1, 2)
The vertex of the positive parabola turns at point (-2, -11)
It is (1, 1).
Easier to show you a simple example as I forget the formulaic approach. X2 + 4X - 6 = 0 add 6 to each side x2 + 4X = 6 Now, halve the linear term ( 4 ), square it and add it to both sides X2 + 4X + 4 = 6 + 4 gather the terms on the right side and factor the left side (X + 2)2 = 10 subtract 10 from each side (X + 2)2 - 10 = 0 (- 2, - 10 ) -------------------the vertex of this quadratic function
-2, 6
3
That would be a horizontal parabola, with it's vertex pointing to the left: y2 = 4x + 6 y2 - 6 = 4x x = y2/4 - 3/2 Now let's find it's vertex, by taking it's derivative and finding the point at which it equals 0: x' = y/2 y = 0 x = 02/4 - 3/2 x = -3/2 So it's vertex occurs at the point (-3/2, 0) Now let's find out where it intercepts the y-axis: x = y2/4 - 3/2 0 = y2/4 - 3/2 y2/4 = 3/2 y2 = 6 y = ±√6 So it intercepts the y-axis at the points (0, -|√6|) and (0, +|√6|)