SOLUTION: Need Help!! Find the point(s) of intersection of the parabola y=x^2+4x+3 and the line y=x+1.

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Question 264901: Need Help!!
Find the point(s) of intersection of the parabola y=x^2+4x+3 and the line
y=x+1.
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Subtract the 2nd equation from the 1st
y+=+x%5E2+%2B+4x+%2B+3
-y+=+-x+-+1
0+=+x%5E2+%2B+3x+%2B+2
Use the quadratic formula to find roots
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
a+=+1
b+=+3
c+=+2
x+=+%28-3+%2B-+sqrt%28+3%5E2-4%2A1%2A2+%29%29%2F%282%2A1%29+
x+=+%28-3+%2B-+sqrt%28+9+-+8+%29%29%2F2+
x+=+%28-3%2B1%29%2F2
x+=+-1
and
x+=+%28-3-1%29%2F2
x+=+-2
The points I have so far are
(-1,?) and (-2,?)
To find y-components, plug x's back into original equatiuon
y+=+x%5E2+%2B+4x+%2B+3
y+=+%28-1%29%5E2+%2B+4%2A%28-1%29+%2B+3
y+=+1+-+4+%2B+3
y+=+0
I have point (-1,0) now
and
y+=+x%5E2+%2B+4x+%2B+3
y+=+%28-2%29%5E2+%2B+4%2A%28-2%29+%2B+3
y+=+4+-+8+%2B+3
y+=+-1
The other point is (-2,-1)
I'll plot them to check answers
+graph%28+500%2C+500%2C+-6%2C+6%2C+-6%2C+6%2C+x%5E2+%2B+4x+%2B+3%2Cx+%2B+1%29+