Question

Use the method of completing the square to write the equations of the given parabola in this form: (y-k)=a(x-h)^2 where a =0, (h,k) is the vertex, and x=h is the axis of symmetry. Find the vertex of this parabola: y=-4x^2-8x-7

Answer 1

Answer:

(-1, -3)

Step-by-step explanation:

$(y-k)=a(x-h)^2,\ a\neq0$

$(a\pm b)^2=a^2\pm2ab+b^2\qquad(*)$

We have

$y=-4x^2-8x-7$

$y=-4x^2-4(2x)-4(1)-3$

distribute

$y=-4(\underbrace{x^2+2x+1}_{(*)})-3$  

use (*)

$y=-4(x+1)^2-3$

add 3 to both sides

$y+3=-4(x+1)^2$

$y-(-3)-4\bigg(x-(-1)\bigg)^2$

the vertex

$(-1;\ -3)$