Answer
The graph is shown below.
Range $[-5,\infty)$.
Work Step by Step
The given function is a quadratic function.
$f(x)=x^2+4x-1$
The standard form of the quadratic function is
$f(x)=ax^2+bx+c$
Compare both equations $a=1,b=4$ and $c=-1$.
Step 1:- Parabola opens.
$a>0$, the parabola open upward.
Step 2:- Vertex.
$x-$coordinate of the vertex is $x=-\frac{b}{2a}=-\frac{(4)}{2(1)}=-2$.
Substitute the value of $x$ into the function.
$\Rightarrow f(-2)=(-2)^2+4(-2)-1$
Simplify.
$\Rightarrow f(-2)=4-8-1$
$\Rightarrow f(-2)=-5$
The vertex is $(-2,-5)$.
Step 3:- $x-$intercepts.
Replace $f(x)$ with $0$ into the given function.
$\Rightarrow 0=x^2+4x-1$
Use quadratic formula, we have $a=1,b=4$ and $c=-1$.
$\Rightarrow x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
Substitute all values.
$\Rightarrow x=\frac{-(4)\pm\sqrt{(4)^2-4(1)(-1)}}{2(1)}$
Simplify.
$\Rightarrow x=\frac{-4\pm\sqrt{16+4}}{2}$
$\Rightarrow x=\frac{-4\pm\sqrt{20}}{2}$
$\Rightarrow x=\frac{-4\pm2\sqrt5}{2}$
Separate the fractions.
$\Rightarrow x=\frac{-4+2\sqrt5}{2}$ or $x= \frac{-4-2\sqrt5}{2}$
Simplify.
$\Rightarrow x=-2+\sqrt5$ or $ x=-2-\sqrt5$
The $x-$intercepts are $-2+\sqrt5$ and $-2-\sqrt5$. The parabola passes through $(-2+\sqrt5,0)$ and $(-2-\sqrt5,0)$.
Step 4:- $y-$intercept.
Replace $x$ with $0$ in the given function.
$\Rightarrow f(0)=(0)^2+4(0)-1$
Simplify.
$\Rightarrow f(0)=-1$
The $y-$intercept is $-1$. The parabola passes through $(0,-1)$.
Step 5:- Graph.
Use the points:- vertex, $x-$intercepts and $y-$intercept to draw a parabola.
The axis of symmetry is $x=-2$.
$A=(-2,-5)$
$B=(-2+\sqrt5,0)$
$C=(-2-\sqrt5,0)$
$D=(0,-1)$.
From the graph the range of the function is
$[-5,\infty)$.
