ANSWER
The correct answer is B
EXPLANATION
The first function is
[tex]x^2-2x+y-3=0[/tex]
We make y the subject to obtain;
[tex]y=-x^2+2x+3[/tex]
Let us quickly write this in the vertex form.
[tex]y=-(x^2-2x)+3[/tex]
[tex]y=-(x^2-2x+(-1)^2)+3+(-1)^2[/tex]
[tex]y=-(x-1)^2+4[/tex]
Since the [tex]a[/tex] is negative, the graph opens up.
The vertex is at [tex](1,4)[/tex]
The y-intercept is [tex]-3[/tex]
The x-intercept is found by equating the function to zero.
[tex]-(x-1)^2+4=0[/tex]
[tex]\Rightarrow (x-1)^2=4[/tex]
[tex]\Rightarrow (x-1)=\pm 2[/tex]
[tex]\Rightarrow x=1 \pm2[/tex]
[tex]\Rightarrow x=-1,3[/tex]
With these information we can quickly sketch the graph as shown in the attachment(the red graph).
For the second function,
[tex]x^2+y=0[/tex]
we again make y the subject to obtain,
[tex]y=-x^2[/tex]
This is a basic quadratic function that can be graphed easily. Note that it is also a maximum graph.
From the graph the solution to the two functions is
[tex](-1.5,-2.25)[/tex]
