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 \[{{x}^{2}}+10x+9\], the coefficient of \[{{x}^{2}}\]is 1.
As, \[a>0\], thus the parabola opens upwards.
(b)
Compare the given equation of the parabola\[y={{x}^{2}}+10x+9\]with the general equation and obtain the values of \[a,b\]and\[c\]as
\[\begin{align}
& a=1 \\
& b=10 \\
& c=9
\end{align}\]
Now, the\[x\]coordinate of the vertex is obtained as,
\[\begin{align}
& x=-\frac{b}{2a} \\
& =-\frac{10}{2\times 1} \\
& =-\frac{10}{2} \\
& =-5
\end{align}\]
Now, substitute above \[x\]coordinate in the equation of the parabola to get the\[y\]coordinate of the vertex as,
\[\begin{align}
& y={{\left( -5 \right)}^{2}}+10\left( -5 \right)+9 \\
& y=25-50+9 \\
& y=-16
\end{align}\]
Thus, the vertex is\[\left( -5,-16 \right)\].
(c)
Substitute\[y=0\]in the parabola equation and solve for\[x\]as,
\[\begin{align}
& {{x}^{2}}+10x+9=0 \\
& {{x}^{2}}+x+9x+9=0 \\
& x\left( x+1 \right)+9\left( x+1 \right)=0 \\
& \left( x+1 \right)\left( x+9 \right)=0
\end{align}\]
Thus, the given parabola cuts the \[x-\text{axis}\]at two points, that is, \[x=-1\]and \[x=-9\].
Hence, the coordinates of \[x-\text{intercepts}\]of the parabola are \[\left( -1,0 \right)\]and \[\left( -9,0 \right)\].
(d)
Substitute\[x=0\]in the parabola equation and solve for\[y\]as
\[\begin{align}
& y={{x}^{2}}+10x+9 \\
& y=0+0+9 \\
& y=9
\end{align}\]
Thus, the given parabola cuts the\[y\]axis at\[y=9\].
Hence, the coordinates of \[y-\text{intercepts}\]of the parabola is \[\left( 0,9 \right)\].
