Given `y=x^2+6x+5` :

The vertex will be the minimum value of this function (the graph is a parabola opening up), and the axis of symmetry is the vertical line through the vertex.

To find the minimum take the first derivative to get `f'(x)=2x+6` .

The minimum will occur where `f'(x)=0` ...

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Given `y=x^2+6x+5` :

The vertex will be the minimum value of this function (the graph is a parabola opening up), and the axis of symmetry is the vertical line through the vertex.

To find the minimum take the first derivative to get `f'(x)=2x+6` .

The minimum will occur where `f'(x)=0` (extrema occur only at critical points; since polynomials are infinitely differentiable the only critical points will occur when the derivative is zero.)

2x+6=0 ==> x=-3

**Thus the vertex is at (-3,f(-3)) or (-3,-4)**

**The axis of symmetry is x=-3**

**The function factors as `y=(x+1)(x+5)` so the x-intercepts are at -1 and -5. The y-intercept is at 0, so f(0)=5 implies the y-intercept is 5.**

**The domain is all real numbers.**

**The range is `y>=-4` as (-3,-4) is an absolute minimum.**

To find the intervals where the function is increasing or decreasing we use the first derivative:

**`2x+6>0 ==> x > -3` so the function increases on `(-3,oo)` **

**2x+6<0 ==> x<-3 so the function decreases on `(-oo,-3)` **

The graph: