Answer
The axis of symmetry is $x=1$.
domain: $(-\infty,\infty)$
range: $[-16,\infty)$
.
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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$f(x)=x^{2}-2x-15$
$f(x)=(x^{2}-2x+1)-15-1$
$f(x)=(x-1)^{2}-16$
1. opens up (a=$1>0$).
2. vertex: $(1, -16). $The axis of symmetry is $x=1.$
3. x-intercepts:
$0=(x-1)^{2}-16$
$(x-1)^{2}=16$
$x-1=\pm 4$
$x=-3$ ,$\quad x=5$
4. y-intercept:
$f(0)=0^{2}-2(0)-15=-15$
5. graph....
