Answer:
(1,-3) and (-2,0)
Step-by-step explanation:
[tex]x+y=-2[/tex]
[tex]y=x^2-4[/tex]
Applying the second equation to the first
[tex]x+x^2-4=-2\\\Rightarrow x+x^2-4+2=0\\\Rightarrow x^2+x-2=0[/tex]
Solving the equation we get
[tex]x=\frac{-1+\sqrt{1^2-4\cdot \:1\left(-2\right)}}{2\cdot \:1}, \frac{-1-\sqrt{1^2-4\cdot \:1\left(-2\right)}}{2\cdot \:1}\\\Rightarrow x=1, -2[/tex]
When x = 1
[tex]y=-2-1\\\Rightarrow y=-3[/tex]
When x = -2
[tex]y=-2-(-2)\\\Rightarrow y=0[/tex]
So, the line and curve will intersect at points (1,-3) and (-2,0)
