PLS HELP a parabola has the equation y = 4x^2 + bx -2 and the equation for the axis of symmetry is x = -1.

Step-by-step explanation:

Remark

Unless you know calculus, the only way to solve this is by completing the square.

Solution

y = 4x^2 + bx - 2

y = 4(x^2 + b/4 ) - 2

y = 4(x^2 + b/4 + (1/2b/4)^2 ) - 2 - 4*(b/8)^2

y = 4(x^2 + b/4 + (b/8)^2 ) -  2 - b^2/16

y = 4(x + b/8)^2 - 2 - b^2/16

Now what you want to happen is to give b/8 a value of 1 so that when x is put in -1 you get 0 for  what is inside the brackets.  

b/8 = 1

b = 8

y = 4(x + 1)^2 - 2 - (8)^2 / 16

y = 4(x + 2)^2 - 2 - 4

y = 4(x + 1)^2 - 6

The vertex is at (-1, - 6)

The axis of symmetry  is x = - 1

Note

This could have been done a lot shorter just by realizing that the number inside the brackets had to be + 1 as in (x + 1). But then you would not have been able to know the minimum value for y which is - 6

. + λrh() e 1 1 2 4 4 4 4 4 3 4 4 4 4 4 + th−1(− s(x)2 dx ω + λrh−1() e 2 1 2 4 4

step-by-step explanation: