Question

Solve the system of equations below by graphing.

Answer 1

ANSWER

The correct answer is B

EXPLANATION

The first function is

$x^2-2x+y-3=0$

We make y the subject to obtain;

$y=-x^2+2x+3$

Let us quickly write this in the vertex form.

$y=-(x^2-2x)+3$

$y=-(x^2-2x+(-1)^2)+3+(-1)^2$

$y=-(x-1)^2+4$

Since the $a$ is negative, the graph opens up.

The vertex is at $(1,4)$

The y-intercept is $-3$

The x-intercept is found by equating the function to zero.

$-(x-1)^2+4=0$

$\Rightarrow (x-1)^2=4$

$\Rightarrow (x-1)=\pm 2$

$\Rightarrow x=1 \pm2$

$\Rightarrow x=-1,3$

With these information we can quickly sketch the graph as shown in the attachment(the red graph).

For the second function,

$x^2+y=0$

we again make y the subject to obtain,

$y=-x^2$

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

$(-1.5,-2.25)$