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 \[f\left( x \right)=-{{x}^{2}}-2x+3\], \[a=-1.\]
The parabola opens downward as the value of a is negative.
Hence, the provided parabola opens downward.
(b)
In the equation \[f\left( x \right)=-{{x}^{2}}-2x+3,\]\[a=-1\], \[b=-2\], and \[c=3.\]
So, x-coordinate is
\[\begin{align}
& x=\frac{-\left( -2 \right)}{2\left( -1 \right)} \\
& =-1
\end{align}\]
Substitute x value in the equation:
\[\begin{align}
& f\left( x \right)=-{{x}^{2}}-2x+3 \\
& =-{{1}^{2}}-2\left( -1 \right)+3 \\
& =4
\end{align}\]
So, the vertex is at \[\left( -1,4 \right).\]
(c)
To find x-intercepts, make\[y=0.\]
\[0=-{{x}^{2}}-2x+3\]
Solve the quadratic equation by factoring.
\[\begin{align}
& {{x}^{2}}+2x-3=0 \\
& {{x}^{2}}+3x-x-3=0 \\
& \left( x+3 \right)\left( x-1 \right)=0
\end{align}\]
So
\[\left( x+3 \right)=0\]or\[\left( x-1 \right)=0\]
\[x=-3\], \[x=1\]
So, the parabola passes through \[\left( 1,0 \right)\] and \[\left( -3,0 \right).\]
Hence,x-intercepts of the parabola are \[\left( 1,0 \right)\]and \[\left( -3,0 \right)\].
(d)
To find y-intercept, make \[x=0\].
\[\begin{align}
& y=-{{\left( 0 \right)}^{2}}-2\left( 0 \right)+3 \\
& =3
\end{align}\]
So
\[y=3\]
The parabola passes through \[\left( 0,3 \right).\]
Hence,y-intercept of the parabola is \[\left( 0,3 \right)\].
