Answer
The graph is shown below.
Range $[2,\infty)$.
Work Step by Step
The given function is a quadratic function:
$f(x)=6-4x+x^2$
Rewrite as.
$f(x)=x^2-4x+6$
The standard form of the quadratic function is
$f(x)=ax^2+bx+c$
Compare both equations $a=1,b=-4$ and $c=6$.
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.
$\Rightarrow f(2)=(2)^2-4(2)+6$
Simplify.
$\Rightarrow f(2)=4-8+6$
Simplify.
$\Rightarrow f(2)=2$
The vertex is $(2,2)$.
Step 3:- $x-$intercepts.
Replace $f(x)$ with $0$ into the given function.
$\Rightarrow 0=x^2-4x+6$
Use quadratic formula, we have $a=1,b=-4$ and $c=6$.
$\Rightarrow x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
Substitute all values.
$\Rightarrow x=\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(6)}}{2(1)}$
Simplify.
$\Rightarrow x=\frac{4\pm\sqrt{16-24}}{2}$
$\Rightarrow x=\frac{4\pm\sqrt{-8}}{2}$
The equation has imaginary solutions, there are no $x-$intercepts.
Step 4:- $y-$intercept.
Replace $x$ with $0$ in the given function.
$\Rightarrow f(0)=(0)^2-4(0)+6$
Simplify.
$\Rightarrow f(0)=6$
The $y-$intercept is $6$. The parabola passes through $(0,6)$.
Step 5:- Graph.
Use the points:- vertex, $x-$intercepts and $y-$intercept to draw a parabola.
The axis of symmetry is $x=2$.
$A=(2,2)$
$B=(0,6)$.
From the graph the range of the function is
$[2,\infty)$.
