Find the points of intersection of the graphs of the equations.
x = 3 − y^2
y = x − 1
from the 2nd :
y = x − 1
x = y+1
sub that into the first:
y+1 = 3-y^2
y^2 + y - 2 = 0
(y+2)(y-1) = 0
so y = -2 or y = 1
if y = -2, x = -2+1 = -1, so they intersect at (-1,-2)
if y = 1, x = 1+1 = 2, and they intersect at (2,1)
verification:
verification:
http://www.wolframalpha.com/input/?i=plot+x+%3D+3+%E2%88%92+y%5E2+,+y+%3D+x+%E2%88%92+1
Thank you I understand now
To find the points of intersection of the graphs of the equations x = 3 - y^2 and y = x - 1, we need to solve these equations simultaneously.
First, let's substitute the expression for x from the second equation into the first equation:
x = 3 - y^2
x = y + 1
Now we can set these two expressions for x equal to each other:
y + 1 = 3 - y^2
Rearrange the equation to bring all terms to one side:
y^2 + y - 2 = 0
Now we can factorize the quadratic equation:
(y + 2)(y - 1) = 0
This gives us two possible solutions for y:
1) y + 2 = 0 --> y = -2
2) y - 1 = 0 --> y = 1
Now we substitute these y-values back into either of the original equations to find the corresponding x-values.
For y = -2:
x = y + 1
x = -2 + 1
x = -1
So, the first point of intersection is (-1, -2).
For y = 1:
x = y + 1
x = 1 + 1
x = 2
Therefore, the second point of intersection is (2, 1).
In conclusion, the graphs of the equations x = 3 - y^2 and y = x - 1 intersect at the points (-1, -2) and (2, 1).