y = x2 + 2x + 8
a = 1 > 0 the parabola opens upward, b = 2, c = 8 the y-intercept is (0, 8).
axis of symmetry: x = -b/2a = -1
The axis of symmetry passes through x = -1, so we have the point (-2, 8) which is symmetrical to (0, 8).
Let f(x) = y, and evaluate f(-1) to find the y-coordinate of the vertex.
f(-1) = (-1)2 + 2(-1) + 8 = 7; vertex = (-1, 7).
Since the parabola opens upward and the vertex lies on the second quadrant, there are no x-intercepts.
Evaluate f(1) = (1)2 + 2(1) + 8 = 11; (1, 11) symmetrical to (-3, 11).
Just plot the points and draw the parabola.
8
X2 + 2X - 14 = 6X2 + 2X - 8 = 0what two products of - 8 = 2 ?(X +4)(X - 2)=========X = - 4X = 2
8
If you mean: y = x^2 -2x -8 then x = -2 or x = 4
x + 7 / x - 2 = x + 4 / x - 3(times both sides by x - 3 and x - 2)(x + 7)(x - 3) = (x + 4)(x - 2)x2 + 7x - 3x - 21 = x2 + 4x - 2x - 8x2 + 4x - 21 = x2 + 2x - 8(- x2 from both sides)4x - 21 = 2x - 8(- 2x from both sides)2x - 21 = -8(+ 21 on both sides)2x = 13(divide both sides by 2)x = 6.5
8
X2 + 2X - 14 = 6X2 + 2X - 8 = 0what two products of - 8 = 2 ?(X +4)(X - 2)=========X = - 4X = 2
10
I think this is how you do it: x2+2x = 8 x2+2x+(1/2*(2))2 = 8 + (1/2*(2))2 x2+2x+1=8+1 (x+1)2 = 9 SQ.ROOT [(x+1)2 = 9] x+1 = 3 and x+1 = -3 so, x=2 and x=-4
2x2 + 2x - 8 = 2(x2 + x - 4). This can take any value greater than or equal to -8.5
8
x2 - 2x + 65 = 0 Subtract 64 from each side: x2 - 2x + 1 = -64 (x - 1)2 = (i*8)2 where i is the imaginary square root of -1 x - 1 = +/- 8i x = 1 +/- 8i
The answer is (3.5)2 plus (4)2 equals 410.
2x+6x=-9 => 8x=-9=> x=-8/9
x2 + 2x - 48 = x2 + 8x - 6x - 48 = x(x + 8) - 6(x + 8) = (x - 6)(x + 8)
-x2 + 2x + 48 = (-x - 6)(x - 8)
x2 + 2x + 2 = 0 Use the quadratic formula: x = (-2 +- sqrt(4 - 8))/2 x = (-2 +- 2i)/2 = -1 +- i