Answer
Axis of symmetry: $ x=1$
Domain: $(-\infty,\infty)$
Range: $(-\infty,\ 4 )$
Work Step by Step
See page 335:
Graphing Quadratic Functions with Equations in the Form
$ f(\mathrm{x})=ax^{2}+bx+c.$
1. Determine whether the parabola opens upward or downward. If $a>0$, it opens upward. If $a<0$, it opens downward.
$a=-1$, opens downward.
2. Determine the vertex of the parabola. The vertex is
$(-\displaystyle \frac{b}{2a}, f(-\frac{b}{2a}))$.
$-\displaystyle \frac{b}{2a}=-\frac{2}{2(-1)}=1,\qquad f(1)=-1^{2}+2(1)+3=4$
Vertex: $(1,4).\quad $Axis of symmetry:$ x=1$
3. Find any x-intercepts by solving $f(x)=0$.
$-x^{2}+2x+3=0\quad/\times(-1)$
$x^{2}-2x-3=0$
$(x-3)(x+1)=0$
$x=-1$, or $x=3$
4. Find the y-intercept by computing $f(0)$.
Because $f(0)=c$ (the constant term in the function's equation), the y-intercept is $c$ and the parabola passes through $(0, c) = (0,3)$
5. Plot the intercepts, the vertex, and additional points as necessary. Connect these points with a smooth curve.
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Axis of symmetry:$ x=1$
Domain: $(-\infty,\infty)$
Range: $(-\infty,\ 4 )$