Question

y=-x^2-6x
interval when function is increasing?
interval when function is decreasing?

Answer 1

The derivative is

\[y \prime = -2x-6\]

When \[y \prime > 0\]
The function is increasing

If \[y \prime < 0\]

The function is decreasing, then?...

Answer 2

The derivative changes in x=-3

So is increasing in the interval\[(-\infty, -3)\]
(Because y'>0 or positive)
and decreasing in
\[(-3,\infty)\]
(because y' is negative)

Answer 3

Thank you.. how to do I fiqure out the vertex

Answer 4

Complete the polynomial, to find the vertex

Answer 5

\[y=-x ^{2}-6x\]
\[y=-(x ^{2}+6x)\]
Rounding out the perfect square trinomial (I used google traductor)
\[y=-\left[ \left( x+3 \right)^{2}-9 \right]\]
\[y=-(x+3)^{2}+9\]
\[y-9=-(x+3)^{2}\]
From this equation we can gen the vertex V(-3,9)