That will be a vertical line that passes through the parabola at the point where it's slope is equal to zero.
The first step then, is to take the derivative of the curve with respect to x:
y = x2 + 2x + 6
dy/dx = 2x + 2
Then let dy/dx equal zero, and solve for x:
0 = 2x + 2
2x = -2
x = -1
So the axis of symmetry is the line x = -1.
y=x2+2x+1 -b -2 2a= 2= -1 = axis of symmetry is negative one.
-1
the line of symmetry would occur at x=0
-x2 + 2x + 48 = (x +6)(8 - x)
x2 + 2x -6 = 0 x2 + 2x + 1 = 7 (x + 1)2 = 7 x = -1 ± √7
It is at: (-1, 0)
y=x2+2x+1 -b -2 2a= 2= -1 = axis of symmetry is negative one.
-1
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)
the line of symmetry would occur at x=0
y = x2 + 2x + 1zeros are:0 = x2 + 2x + 10 = (x + 1)(x + 1)0 = (x + 1)2x = -1So that the graph of the function y = x + 2x + 1 touches the x-axis at x = -1.
Line of symmetry: x = 3
y = x2 + 8x - 7 a = 1, b = 8, c = -7 the equation of the axis of symmetry: x = -b/2a x = -8/(2*1) = -4
-x2 + 2x + 48 = (x +6)(8 - x)
x2 + 2x -6 = 0 x2 + 2x + 1 = 7 (x + 1)2 = 7 x = -1 ± √7
-x2 + 2x + 48 = (-x - 6)(x - 8)
x2+2x-63 = (x-7)(x+9) when factored