2x2 + 4 + 1 = 2x2 + 5 So, the vertex is (0, 5)
The vertex has a minimum value of (-4, -11)
We need to complete the square to find the vertex of this parabola. X^2 + 2X + 5 = 0 X^2 + 2X = -5 halve the coefficient of the linear term ( 2X ) and square it perfectly and add to other side (X + 1)^2 = -5 + 1 (X + 1)^2 = -4 (X + 1)^2 + 4 = 0 Vertex is.... X = -1 Y = 4
-2-5
If you are using a calculator just plug it in and hit graph. If you are doing it by hand, start with making a X-Y Table. Plug in X values into the equation to get a Y value out. Plot about 5 points on the graph to get a basic look at the parabola. To get the right the values, you want to start with the vertex and go out from there. To start, you need to find the axis of symmetry (-b/2a) [From the basic equation of ax squared +bx + c] That is the X Value for the vertex. Plug that in to find the Y Value for the vertex. The more points you find the more accurate the graph but normally 5 is enough (vertex and two on left and right)
2x2 + 4 + 1 = 2x2 + 5 So, the vertex is (0, 5)
The vertex has a minimum value of (-4, -11)
We need to complete the square to find the vertex of this parabola. X^2 + 2X + 5 = 0 X^2 + 2X = -5 halve the coefficient of the linear term ( 2X ) and square it perfectly and add to other side (X + 1)^2 = -5 + 1 (X + 1)^2 = -4 (X + 1)^2 + 4 = 0 Vertex is.... X = -1 Y = 4
plot 4,-5 2,2 and 7,-5 on a graph than connect the vertex
y=-2x^2+8x+3
-2-5
If you are using a calculator just plug it in and hit graph. If you are doing it by hand, start with making a X-Y Table. Plug in X values into the equation to get a Y value out. Plot about 5 points on the graph to get a basic look at the parabola. To get the right the values, you want to start with the vertex and go out from there. To start, you need to find the axis of symmetry (-b/2a) [From the basic equation of ax squared +bx + c] That is the X Value for the vertex. Plug that in to find the Y Value for the vertex. The more points you find the more accurate the graph but normally 5 is enough (vertex and two on left and right)
No, the complete graph of 5 vertices is non planar. because we cant make any such complete graph which draw without cross over the edges . if there exist any crossing with respect to edges then the graph is non planar.Note:- a graph which contain minimum one edge from one vertex to another is called as complete graph...
You can do the equation Y 2x plus 3 on a graph. On this graph the Y would equal 5 and X would equal to 0.
Factorising, we have, 9x2+ 30x + 25 = (3x + 5)2;thus, when y = 0, x = -1⅔. This tells us that the sole x-intercept of this function is at the vertex, (0, 1⅔).
x = -3y = -14
(-3, -5)