How would i solve the system fo equations? is there any way i could get the points of the equation?

(1,3) or (3,7)

Step-by-step explanation:

y=x²-2x+4 i

y=2x+1 ii

Equating i and ii,

x²-2x+4=2x+1

x²-2x-2x+4-1=0

x²-4x+3=0

x²-x-3x+3=0

x(x-1)-3(x-1)=0

(x-1)(x-3)=0.

Either, x=1 or x=3.

If x=1,y=3 and if x=3, y=7

The points are (1,3) or (3,7)

step-by-step explanation:

through just a quick inspection, one can see that the graph of this cubing function starts in quadrant iii and ends in quadrant i.   when x = 0, y = 3, so the y-intercept is (0, 3).

if graphs are part of the problem statement, share them.   .

where is the page?

step-by-step explanation: