What is the vertex of x^2+x-5=0

The vertex of parabola ax

Your parabola has a=1, b=2, so the vertex is located at x=-2/2 = -1

y = (-1)

The coordinates of the vertex are (-1, -6)

(Look at the picture.)

^{2}+bx+c is located at x=-b/(2a).Your parabola has a=1, b=2, so the vertex is located at x=-2/2 = -1

y = (-1)

^{2}+ 2(-1) -5 = 1 -2 -5 = -6The coordinates of the vertex are (-1, -6)

(Look at the picture.)

Y = 6(x+2)^2 - 4

Y = x^2+2x-5....(1)

The vertex of parabola for ax^2+bx+c is x = -b/2a

so, a=1, b=2, c=-5

x = -2/(2*1)

x = -2/2 = -1

put the value of x into eq. (1)

y = (-1)^2+2(-1)-5

y = 1-2-5

y = -6

so, points for vertex of parabola are : (-1, -6)

The vertex of parabola for ax^2+bx+c is x = -b/2a

so, a=1, b=2, c=-5

x = -2/(2*1)

x = -2/2 = -1

put the value of x into eq. (1)

y = (-1)^2+2(-1)-5

y = 1-2-5

y = -6

so, points for vertex of parabola are : (-1, -6)

Y=15x2+7x-2