Answer
See below:
Work Step by Step
(a)
The equation of parabola is\[y=a{{x}^{2}}+bx+c\]. The coefficient a,determines the opening of the parabola.
In \[y={{x}^{2}}-6x-7\], \[a=1\].
Parabola opens upwards as the value of a is positive.
(b)
Since, thex-coordinate of vertex of a parabola \[y=a{{x}^{2}}+bx+c\] is provided by:
\[x=\frac{-b}{2a}\]
In the equation \[y={{x}^{2}}-6x-7\]:
\[a=1\], \[b=-6\], \[c=-7\]
So, x-coordinate is:
\[\begin{align}
& x=\frac{-\left( -6 \right)}{2\left( 1 \right)} \\
& =3
\end{align}\]
Substituting x value in the equation:
\[\begin{align}
& y={{x}^{2}}-6x-7 \\
& ={{3}^{2}}-6\left( 3 \right)-7 \\
& =-16
\end{align}\]
So, the vertex is at \[\left( 3,-16 \right)\].
(c)
To find x-intercepts,
make\[y=0\], \[0={{x}^{2}}-6x-7\] and solve the quadratic equation by factoring,
\[\begin{align}
& {{x}^{2}}-6x-7=0 \\
& {{x}^{2}}-7x+x-7=0 \\
& \left( x-7 \right)\left( x+1 \right)=0
\end{align}\]
So,
\[\left( x-7 \right)=0\]or\[\left( x+1 \right)=0\]
\[x=7\], \[x=-1\]
The parabola passes through \[\left( -1,0 \right)\]and \[\left( 7,0 \right)\].
(d)
To find y-intercept, make \[x=0\]:
\[\begin{align}
& y={{\left( 0 \right)}^{2}}-6\left( 0 \right)-7 \\
& =-7
\end{align}\]
So,
\[y=-7\]
The parabola passes through \[\left( 0,-7 \right)\].
