Answer
Standard form: $f(x)=(x+\frac{5}{2})^2-\frac{41}{4}$
Vertex: $(-\frac{5}{2},-\frac{41}{4})$
Axis of symmetry: $x=-\frac{5}{2}$.
x-intercepts: $(\frac{-5-\sqrt {41}}{2},0)$ and $(\frac{-5+\sqrt {41}}{2},0)$
Work Step by Step
$f(x)=x^2+5x-4~~$ ($a=1,~b=5,~c=-4$)
$f(x)=x^2+5x+\frac{25}{4}-\frac{25}{4}-4$
$f(x)=[x^2+2(\frac{5}{2})x+(\frac{5}{2})^2]-\frac{41}{4}$
$f(x)=(x+\frac{5}{2})^2-\frac{41}{4}$
$-\frac{b}{2a}=-\frac{5}{2(1)}=-\frac{5}{2}$
$f(-\frac{5}{2})=(-\frac{5}{2}+\frac{5}{2})^2-\frac{41}{4}=-\frac{41}{4}$
Vertex: $(-\frac{b}{2a},f(-\frac{b}{2a}))=(-\frac{5}{2},-\frac{41}{4})$
So, the axis of symmetry is $x=-\frac{b}{2a}=-\frac{5}{2}$.
The x-intercepts:
$f(x)=0$
$(x+\frac{5}{2})^2-\frac{41}{4}=0$
$(x+\frac{5}{2})^2=\frac{41}{4}$
$x+\frac{5}{2}=±\frac{\sqrt {41}}{2}$
$x=-\frac{5}{2}±\frac{\sqrt {41}}{2}=\frac{-5±\sqrt {41}}{2}$
So, the x-intercept points are: $(\frac{-5-\sqrt {41}}{2},0)$ and $(\frac{-5+\sqrt {41}}{2},0)$
