The vertex is at (-1,0).
y = x +1 is the equation of a straight line and so has no vertex.
y = x2 + 14x + 21 a = 1, b = 14 x = -b/2a = -14/2*1 = -7
Need signs between terms.
The vertex coordinate point of the vertex of the parabola y = 24-6x-3x^2 when plotted on the Cartesian plane is at (-1, 27) which can also be found by completing the square.
The vertex is at (-1,0).
Interpreting that function as y=x2+2x+1, the graph of this function would be a parabola that opens upward. It would be equivalent to y=(x+1)2. Its vertex would be at (-1,0) and this vertex would be the parabola's only zero.
3
y = x +1 is the equation of a straight line and so has no vertex.
y = x2 + 14x + 21 a = 1, b = 14 x = -b/2a = -14/2*1 = -7
Need signs between terms.
The vertex coordinate point of the vertex of the parabola y = 24-6x-3x^2 when plotted on the Cartesian plane is at (-1, 27) which can also be found by completing the square.
y = 3x2+2x-1 Line of symmetry: x = -1/3 Vertex coordinate: (-1/3, -4/3)
Take the derivative: y' = 2x + 4. Set equal to zero: 2x + 4 = 0 --> x = -2. Substitute into original and y = -3. Axis of symmetry is x = -2. Vertex is (-2,-3)
1 is a number. It does not have a vertex!
If you meant, your two equations are y=2x+2 and y= 8x+1 your vertex will be at x=1/6 and y=7/3 you can plug these in and they will check out. but if you meant, y=(2x2) +8x+1 y=4+8x+1 y=8x+5 and then you only have one line so there will be no vertex. but the y intercept is at (0, 5) and your x-intercept at (-5/8, 0) if neither are these are what you are saying, sorry, and maybe try and repost and I will help you out.
A vertex !