SOLUTION: Find where two parabolas y= -2(x-3)^2 + 4 and y=2(x-2)^2 -1 intersect each other. I keep coming up with another quadratic equation instead of a point! I also tried sketching bot

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Find where two parabolas y= -2(x-3)^2 + 4 and y=2(x-2)^2 -1 intersect each other. I keep coming up with another quadratic equation instead of a point! I also tried sketching bot      Log On


   

Question 320390: Find where two parabolas y= -2(x-3)^2 + 4 and y=2(x-2)^2 -1 intersect each other.
I keep coming up with another quadratic equation instead of a point! I also tried sketching both parabolas, but the exact point of intersection is difficult to verify.
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
You're getting a quadratic equation because they intersect at two points.
If they only intersected at one point, you would get a linear equation.
.
.

.
.
.
-2%28x-3%29%5E2%2B4=2%28x-2%29%5E2+-1
-2%28x%5E2-6x%2B9%29%2B4=2%28x%5E2-4x%2B4%29-1
-2x%5E2%2B12x-18%2B4=2x%5E2-8x%2B8-1
4x%5E2-20x%2B21=0
Factor
%282x-7%29%282x-3%29=0
Two solutions:
2x-7=0
2x=7
x=7%2F2
Then use either equation to find y.
y=2%287%2F2-4%2F2%29%5E2-1
y=2%283%2F2%29%5E2-1
y=9%2F2-2%2F2
y=7%2F2
.
.
.
2x-3=0
2x=3
x=3%2F2
Again, find y,
y=2%283%2F2-4%2F2%29%5E2-1
y=2%28-1%2F2%29%5E2-1
y=-1%2F2-2%2F2
y=-1%2F2
.
.
.

.
.
.
(7/2,7/2) and (3/2,-1/2)