Answer
The axis of symmetry is $x=2.$
domain: $(-\infty,\infty)$
range: $[2,\ \infty)$
.
Work Step by Step
First, rewrite $f(x)=ax^{2}+bx+c$ in standard form, $f(x)=a(x-h)^{2}+k$
Graphing:
1. Determine whether the parabola opens upward or downward.
If $a>0$, it opens upward. If $a<0$, it opens downward.
2. Determine the vertex of the parabola. The vertex is $(h, k)$.
3. Find any x-intercepts by solving $f(x)=0$.
The function's real zeros are the x-intercepts.
4. Find the y-intercept by computing $f(0)$.
5. Plot the intercepts, the vertex, and additional points as necessary
Connect these points with a smooth curve that is shaped like a bowl or an inverted bowl.
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$f(x)=6-4x+x^{2}$
$f(x)=x^{2}-4x+6$
$f(x)=(x^{2}-4x+4-4)+6$
$f(x)=(x-2)^{2}+2$
1. opens up (a = $1 > 0$).
2. vertex: $(2, 2). $The axis of symmetry is $x=2.$
3. x-intercepts:
$0=(x-2)^{2}+2$
$(x-2)^{2}=-2$
(no real solutions, square can not be negative)
No x-intercepts.
4. y-intercept:
$f(0)=6-4(0)+(0)^{2}=6$
The axis of symmetry is $x=2.$
domain: $(-\infty,\infty)$
range: $[2,\ \infty)$
