Question

Graph the functions on the same coordinate axis. {f(x)=−2x+1g(x)=x2−2x−3

What are the solutions to the system of equations?

select each answer




(2, 3)

(−2, 5)

(2, −3)

(2, 5)

(−2, −3)

Answer 1

so you dont have to suffer lol heres a screenshot

Explanation:

Answer 2

(2,-3) and (-2,5)

Explanation:

Let us graph the two equations one by one.

1.
`f(x)=-2x+1`

If we compare this equation with the slope intercept form of a line which is given as

`y=mx+c`

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`

`y=-2+1 = -1`

Let us find another point by putting x = 2 and solving it for y

`y=-2(2)+1`

`y=-4+1 = -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`

`y=+1 = 5`

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`

`y=x^2-2x+1-1-3`

`y=(x-1)^2-4`

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

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 ;
`y=y=(0)^2-2(0)-3=0-0-3=-3` ; (0,-3)

ii)x=-1 ;
`y=(-1)^2-2(-1)-3=1+2-3=0` ; (-1,0)

iii) x=2 ;
`y=(2)^2-2(2)-3 = 4-4-3 =-3` ;(2,-3)

iii) x=1 ;
`y=(1)^2-2(1)-3 = 1-2-3 =-4` ;(1,-4)

iii) x=-2 ;
`y=(-2)^2-2(-2)-3 = 4+4-3 =5` ;(-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.