Answer
See below:
Work Step by Step
(a)
General equation of a parabola on \[y-\text{axis}\]is \[y=a{{x}^{2}}+bx+c\].
Now, for the parabola \[-{{x}^{2}}+2x+3\], coefficient of \[{{x}^{2}}\]is \[-1\].
As, \[a<0\], thus the parabola opens downwards.
(b)
Compare the given equation of parabola\[y=-{{x}^{2}}+2x+3\]with the general equation and obtain the values of \[a,b\]and\[c\]as,
\[\begin{align}
& a=-1 \\
& b=2 \\
& c=3
\end{align}\]
Now,\[x\]coordinate of the vertex is obtained as,
\[\begin{align}
& x=-\frac{b}{2a} \\
& =-\frac{2}{2\times \left( -1 \right)} \\
& =\frac{2}{2} \\
& =1
\end{align}\]
Now, substitute above \[x\]coordinate in the equation of the parabola to get the\[y\]coordinate of the vertex as,
\[\begin{align}
& y=-{{\left( 1 \right)}^{2}}+2\left( 1 \right)+3 \\
& y=-1+2+3 \\
& y=4
\end{align}\]
Thus, vertex is\[\left( 1,4 \right)\].
(c)
Substitute\[y=0\]in the parabola equation and solve for\[x\]as,
\[\begin{align}
& -{{x}^{2}}+2x+3=0 \\
& {{x}^{2}}-2x-3=0 \\
& {{x}^{2}}+x-3x-3=0 \\
& x\left( x+1 \right)-3\left( x+1 \right)=0
\end{align}\]
Solve ahead to get the factors as,
\[\left( x-3 \right)\left( x+1 \right)=0\]
Thus, the given parabola cuts the \[x-\text{axis}\]at two points, that is, \[x=3\]and \[x=-1\].
(d)
Substitute\[x=0\]in the parabola equation and solve for\[y\]as,
\[\begin{align}
& y=-{{x}^{2}}+2x+3 \\
& y=0+0+3 \\
& y=3
\end{align}\]
Thus, the given parabola cuts the\[y\]axis at\[y=3\].
