This is a Systems of Equations problem, I'm guessing.
6) To find the point of intersection, I have to first set the two equations together:
[tex]2x+12=-x^2+10x+5[/tex]
I can factor one side:
[tex]2x+12 = -1(x^2-10x-5)[/tex]
I can now solve for x to get the x-coordinate on the point of intersection (I just divided both sides by -1):
[tex]-2x-12 = x^2-10x-5[/tex]
Solve for x:
[tex]-2x-7=x^2-10x[/tex]
[tex]8x-7=x^2[/tex]
Since it's a quadratic, I'll just use the quadratic equation to find the two x's. Remember, it says "find all break points" so there's more than 1:
[tex]0 = x^2-8x+7[/tex]
[tex]x = \frac{-b+ \sqrt{b^2-4ac} }{2a} [/tex]
and
[tex]x = \frac{-b- \sqrt{b^2-4ac} }{2a} [/tex]
Plug in:
[tex]x = \frac{8+ \sqrt{(-8)^2-4(1)(7)} }{2(1)} = \frac{-8+ \sqrt{64-28}}{2} [/tex]
and
[tex]x = \frac{8- \sqrt{(-8)^2-4(1)(7)} }{2(1)} = \frac{-8+ \sqrt{64-28}}{2} [/tex]
Solve both:
[tex]x = \frac{8+ \sqrt{36} }{2} = \frac{8+6}{2}=\frac{14}{2}=7[/tex]
and
[tex]x = \frac{8- \sqrt{36} }{2} = \frac{8-6}{2} = 1 [/tex]
Now that we have the x-coordinates, we can plug both of them in into the equation that we had for y to find the point in form (x, y). So, let's do that, and since we can choose either equation, I'm going to choose the first one because it's much easier. Let's do it:
[tex]y = 2x+12[/tex]
[tex]y = 2(7)+12 = 14+12=26[/tex]
and
[tex]y = 2x+12
[/tex]
[tex]y = 2(1)+12 = 2+12 = 14[/tex]
So, your two (and all of them) points of intersection are (7, 46) and (1, 14).
If any step doesn't make sense, hmu. I'm not going to do the other one because I don't have enough time. Maybe I can help you tomorrow in the DMs.
