Answer:
(2,-3) and (-2,5)
Step-by-step explanation:
Let us graph the two equations one by one.
1. ![f(x)=-2x+1](https://tex.z-dn.net/?f=f%28x%29%3D-2x%2B1)
If we compare this equation with the slope intercept form of a line which is given as
![y=mx+c](https://tex.z-dn.net/?f=y%3Dmx%2Bc)
we see that m = -1 and c =1
Hence the slope of the line is -2 and the y intercept is 1. Hence one point through which it is passing is (0,1) .
Let us find another point by putting x = 1 and solving it for y
![y=-2(1)+1](https://tex.z-dn.net/?f=y%3D-2%281%29%2B1)
![y=-2+1 = -1](https://tex.z-dn.net/?f=y%3D-2%2B1%20%3D%20-1)
Let us find another point by putting x = 2 and solving it for y
![y=-2(2)+1](https://tex.z-dn.net/?f=y%3D-2%282%29%2B1)
![y=-4+1 = -3](https://tex.z-dn.net/?f=y%3D-4%2B1%20%3D%20-3)
Hence the another point will be (2,-3)
Let us find another point by putting x = -2 and solving it for y
![y=-2(-2)+1](https://tex.z-dn.net/?f=y%3D-2%28-2%29%2B1)
![y=+1 = 5](https://tex.z-dn.net/?f=y%3D%2B1%20%3D%205)
Hence the another point will be (-2,5)
Now we have two points (0,1) ,(1,-1) , (2,-3) and (-2,5) we joint them on line to obtain our line
2.
![g(x)=y=x^2-2x-3](https://tex.z-dn.net/?f=g%28x%29%3Dy%3Dx%5E2-2x-3)
![y=x^2-2x+1-1-3](https://tex.z-dn.net/?f=y%3Dx%5E2-2x%2B1-1-3)
![y=(x-1)^2-4](https://tex.z-dn.net/?f=y%3D%28x-1%29%5E2-4)
![(y+4)=(x-1)^2](https://tex.z-dn.net/?f=%28y%2B4%29%3D%28x-1%29%5E2)
It represents the parabola opening upward with vertices (1,-4)
Let us mark few coordinates so that we may graph the parabola.
i) x=0 ;
; (0,-3)
ii)x=-1 ;
; (-1,0)
iii) x=2 ;
;(2,-3)
iii) x=1 ;
;(1,-4)
iii) x=-2 ;
;(-2,5)
Now we plot them on coordinate axis and line them to form our parabola
When we plot them we see that we have two coordinates (2,-3) and (-2,5) are common , on which our graphs are intersecting. These coordinates are solution to the two graphs.