How do you find the vertex of $f(x)= -x^2-6x-5$ ?

Question

1890 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-02T00:54:34

First, find the x value of the vertex with the following formula:

$x_v=(-b)/(2a)$

In your function, the values are:

$a=-1$
$b=-6$
$c=-5$

Plugging those values:
$x_v=(-(-6))/(2(-1))=6/(-2)=-3$

Now you do $f(x_v)$ , which is equal to the y value of the vertex ( $y_v$ ):
$f(x_v)=-(-3)^2-6(-3)-5$
$f(x_v)=-9-(-18)-5$
$f(x_v)=4$

Your vertex is
$(x_v,y_v)$
$(-3,4)$

How do you find the vertex of f(x)= -x^2-6x-5 f(x)= -x^2-6x-5?