I do not understand question #4 pg 156 advanced systems of equations?

Here’s the question, which is a no-calc grid-in.

You weren’t very specific about what you don’t understand, so I’ll just work through it.

I think the main thing you need to recognize here is that you’re given two parabolas in vertex form. The first one has a vertex at (2, 4) and opens up, and the fact that it’s y\geq means any value inside the parabola satisfies. The second one has a vertex at (2, 6) and opens down, and again, the y\leq means any value inside the parabola satisfies.

Because this is a no-calc question, it’s really helpful if you’re able to visualize this from the equations.

Both parabolas center on the z=2 line, and the solution set is the overlap between y=4 and y=6. You’re looking for the smallest x-value inside that overlap. You can set the parabolas equal to each other to find the intersections.

(x-2)^2+4=-(x-2)^2+6\\(x^2-4x+4)+4=-(x^2-4x+4)+6\\x^2-4x+8=-x^2+4x+2\\2x^2-8x+6=0\\x^2-4x+3=0\\(x-3)(x-1)=0

That means the intersections are at x=1 and x=3. Because 1 is the lesser value, that’s the answer.

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