Answer
See below:
Work Step by Step
(a)
The general equation of a parabola on \[y-\text{axis}\]is \[y=a{{x}^{2}}+bx+c\].
Now, for the parabola \[{{x}^{2}}-2x-8\], the coefficient of \[{{x}^{2}}\]is 1.
As, \[a>0\], thus the parabola opens upwards.
(b)
Compare the given equation of the parabola\[y={{x}^{2}}-2x-8\]with the general equation and obtain the values of \[a,b\]and\[c\]as
\[\begin{align}
& a=1 \\
& b=-2 \\
& c=-8
\end{align}\]
Now,\[x\]the coordinate of the vertex is obtained as,
\[\begin{align}
& x=-\frac{b}{2a} \\
& =-\frac{\left( -2 \right)}{2\times 1} \\
& =\frac{2}{2} \\
& =1
\end{align}\]
Now, substitute above \[x\]coordinate in the equation of the parabola to get the\[y\]coordinate of the vertex as,
\[\begin{align}
& y={{\left( 1 \right)}^{2}}-2\left( 1 \right)-8 \\
& y=1-2-8 \\
& y=-9
\end{align}\]
Thus, the vertex is\[\left( 1,-9 \right)\].
(c)
Substitute\[y=0\]in the parabola equation and solve for\[x\]as
\[\begin{align}
& {{x}^{2}}-2x-8=0 \\
& {{x}^{2}}+2x-4x-8=0 \\
& x\left( x+2 \right)-4\left( x+2 \right)=0 \\
& \left( x+2 \right)\left( x-4 \right)=0
\end{align}\]
Thus, the given parabola cuts the \[x-\text{axis}\]at two points, that is, \[x=-2\]and \[x=4\].
(d)
Substitute\[x=0\]in the parabola equation and solve for\[y\]as
\[\begin{align}
& y={{x}^{2}}-2x-8 \\
& y=0+0-8 \\
& y=-8
\end{align}\]
Thus, the given parabola cuts the\[y\]axis at\[y=-8\].
