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 \[{{x}^{2}}+4x-5\], coefficient of \[{{x}^{2}}\]is 1.
As, \[a>0\], thus the parabola opens upwards.
(b)
Compare the given equation of parabola\[y={{x}^{2}}+4x-5\]with the general equation and obtain the values of \[a,b\]and\[c\]as,
\[\begin{align}
& a=1 \\
& b=4 \\
& c=-5
\end{align}\]
Now,\[x\]coordinate of the vertex is obtained as,
\[\begin{align}
& x=-\frac{b}{2a} \\
& =-\frac{4}{2\times 1} \\
& =\frac{-4}{2} \\
& =-2
\end{align}\]
Now, substitute above \[x\]coordinate in the equation of the parabola to get the\[y\]coordinate of the vertex as,
\[\begin{align}
& y={{\left( -2 \right)}^{2}}+4\left( -2 \right)-5 \\
& y=4-8-5 \\
& y=-9
\end{align}\]
Thus, vertex is\[\left( -2,-9 \right)\].
(c)
Substitute\[y=0\]in the parabola equation and solve for\[x\]as,
\[\begin{align}
& {{x}^{2}}+4x-5=0 \\
& {{x}^{2}}-x+5x-5=0 \\
& x\left( x-1 \right)+5\left( x-1 \right)=0 \\
& \left( x+5 \right)\left( x-1 \right)=0
\end{align}\]
Thus, the given parabola cuts the \[x-\text{axis}\]at two points, that is, \[x=-5\]and \[x=1\].
(d)
Substitute\[x=0\]in the parabola equation and solve for\[y\]as,
\[\begin{align}
& y={{x}^{2}}+4x-5 \\
& y=0+0-5 \\
& y=-5
\end{align}\]
Thus, the given parabola cuts the\[y\]axis at\[y=-5\].
