What is the vertex of $y = - {(x + 2)}^{2} - 2 x^{2} - x - 4$ $y= -(x+2)^2-2x^2-x-4$ ?

Question

1368 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-30T08:31:43

Vertex is $(- \frac{5}{6}, - \frac{71}{12})$ $(-5/6,-71/12)$

$y = - {(x + 2)}^{2} - 2 x^{2} - x - 4$ $y=-(x+2)^2-2x^2-x-4$

= $- (x^{2} + 4 x + 4) - 2 x^{2} - x - 4$ $-(x^2+4x+4)-2x^2-x-4$

= $- 3 x^{2} - 5 x - 8$ $-3x^2-5x-8$

= $- 3 (x^{2} + \frac{5}{3} x + {(\frac{5}{6})}^{2}) - (- 3) {(\frac{5}{6})}^{2} - 8$ $-3(x^2+5/3x+(5/6)^2)-(-3)(5/6)^2-8$

= $- 3 {(x + \frac{5}{6})}^{2} + \frac{25}{12} - 8$ $-3(x+5/6)^2+25/12-8$

= $- 3 {(x + \frac{5}{6})}^{2} - \frac{71}{12}$ $-3(x+5/6)^2-71/12$

Now it is in vertex form $y = a {(x - h)}^{2} + k$ $y=a(x-h)^2+k$

and vertex is $(- \frac{5}{6}, - \frac{71}{12})$ $(-5/6,-71/12)$

graph{-(x+2)^2-2x^2-x-4 [-6.876, 3.124, -8.7, -3.7]}

What is the vertex of y=−(x+2)2−2x2−x−4 y= -(x+2)^2-2x^2-x-4?