y=aX2+bX+c ---> y=X2-6X+5vertex formula is -b/2a( b=-6,a=1)vertex=-(-6)/2*1=6/2=3 y=32-6(3)+5---->y=-4so vertex is (3,4)
The equation is linear and so has no vertex.
(3, -21)
y = x2 + 6x + 7a = 1, b = 6, c = 7Since a is positive, the graph opens upward. You can find the vertex coordinates (-b/2a, f(-b/2a)) = (-3, f(-3)) = (-3, -2), so the equation of the axis of symmetry is x = -b/2a = -3, plot the y-intercept point (0, 7), plot the point (-b/a, 7) = (-6, 7), and draw the graph that passes through these points.Or complete the square.y = x2 + 6x + 7y = x2 + 6x + 9 - 9 + 7y = (x2 + 6x + 9) - 2y = (x + 3)2 - 2So start with the graph of y = x2 whose vertex is at the origin. Move it 3 units to the left, and 2 units do
8
The vertex of the graph Y 3 X-12 plus 2 would be -1/3 and -4/3. This is taught in math.
The vertex of the graph Y 3 X-12 plus 2 would be -1/3 and -4/3. This is taught in math.
y=aX2+bX+c ---> y=X2-6X+5vertex formula is -b/2a( b=-6,a=1)vertex=-(-6)/2*1=6/2=3 y=32-6(3)+5---->y=-4so vertex is (3,4)
y=-2x^2+8x+3
The equation is linear and so has no vertex.
one vertex: 3 two vertices: 6 three vertices: 8 total 17
Nothing. Vertices is the plural term from vertex. A vertex is a point where two or more lines meet. They usually terminate (end) there but don't have to. In a 2- or 3-dimensional shape a vertex is a corner.
y = 2x2 + 3x + 6 Since a > 0 (a = 2, b = 3, c = 6) the graph opens upward. The coordinates of the vertex are (-b/2a, f(-b/2a)) = (- 0.75, 4.875). The equation of the axis of symmetry is x = -0.75.
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.
A dual graph is constructed by taking the original graph, which must be planar (no crossing edges) and creating a vertex inside each face of the graph. A face is an enclosed area in the graph and the space outside of the graph is also a face. Once you have created a vertex in every space, you must connect every vertex by crossing each edge in the original graph. For example, a simple triangle is planar and has two faces, one inside and one outside. We would form a vertex inside the triangle and somewhere outside of the triangle. Now, we have three edges we must cross, so starting at the inner vertex, draw three lines with one exiting through exactly 1 side each. You should now have a vertex with 3 lines that exist outside of the triangle. Without crossing them, just simply connect them to the vertex on the outside. This will create a dual of the triangle. It should resemble two vertices connected with three edges. Note that this dual graph is not planar like the original.
x2 + y - 49 = 0y = -x2 + 49First, plot the graph of y = -x2, with a vertex (0, 0), then translate it 49 units up. The vertex becomes (0, 49), which is a maximum point (the parabola opens downward).Or make a table to obtain several corresponding y-values for x = -3, -2, -1, 0, 1, 2, 3. Plot the points (x, y), and draw the graph of y = -x + 49.
(3, -21)