Answer
a) See graph
b) Domain: $(-\infty,\infty)$
Range: $(2,\infty)$
c) Decreasing: $(-\infty,2)$
Increasing: $(2,\infty)$
Work Step by Step
We are given the function:
$f(x)=(x-2)^2+2$
a) Rewrite the function in the form $y=ax^2+bx+c$:
$y=x^2-4x+4+2=x^2-4x+6$
Identify $a,b,c$:
$a=1$
$b=-4$
$c=6$
Because $a>0$, the graph opens up.
Determine the vertex:
$x_V=-\dfrac{b}{2a}=-\dfrac{-4}{2(1)}=2$
$y_V=(2-2)^2+2=2$
$V(x_V,y_V)=(2,2)$
The axis of symmetry is:
$x=x_V\Rightarrow x=2$
Determine the intercepts:
$x=0\Rightarrow y=0^2-4(0)+6=6$
$y=0\Rightarrow (x-2)^2+2=0\Rightarrow (x-2)^2=-2$
There are no $x$-intercepts!!!
Graph the function.
b) The domain of the function is:
$(-\infty,\infty)$
The range is:
$(2,\infty)$
c) The function is decreasing on the interval:
$(-\infty,2)$
The function is increasing on the interval:
$(2,\infty)$
