Answer
The axis of symmetry is $x=-2.$
domain: $(-\infty,\infty)$
range: $(-\infty,9]$
.
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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$f(x)=5-4x-x^{2}$
$f(x)=-x^{2}-4x+5$
$f(x)=-(x^{2}+4x+4-4)+5$
$f(x)=-(x^{2}+4x+4)+4+5$
$f(x)=-(x+2)^{2}+9$
1. opens down (a = $-1 < 0$).
2. vertex: $(-2,9). $The axis of symmetry is $x=-2.$
3. x-intercepts:
$0=-(x+2)^{2}+9$
$(x+2)^{2}=9$
$x+2=\pm 3$
$x=-2\pm 3$
$x=-5$ ,$\quad x=1$
4. y-intercept:
$f(0)=5-4(0)-(0)^{2}=5$
The axis of symmetry is $x=-2.$
domain: $(-\infty,\infty)$
range: $(-\infty,9]$
