Answer:
AB=4
Step-by-step explanation:
1. Since you are finding the intersection points of two parabolas:
a. y=-x²+9
b. y=2x²-3
2. You have to set them equal to each other:
2x²-3= - x²+9
2x²+x=9+3
3x²=12
x²=4
This is the crucial part; the absolute value of x is equal to plus minus the square root of 4, since either -2 squared with parentheses or 2 squared is equal to 4.
√x²=±√4
x=±2
or
x=2; x=-2
3. Then you substitute them into each equation. For this step, any sign 2 will work.
a. y=-(2)²+9
y=-4+9
y=5
b. y=2(2)²-3
y=8-3
y=5
4. So our coordinates will be (2,5) and (-2,5). These are the points of intersection.
5. Now we use the distance formula:
[tex]\sqrt{(x₂-x₁)^{2} +(y₂-y₁)^{2} }[/tex]
The subscripts didn't work for this but I mean the square root of x 2 - x 1 in parentheses plus y 2 -y 1.
[tex]\sqrt{(2+2)^{2} +(5-5)^{2} }[/tex]=
√16=
4
The absolute value rule that I mentioned above doesn't work for this because its a distance and you can't have a negative distance.
So AB=4
