How can i find the intercept points of these two equations (semi-circle and exponential):
y=\(\sqrt{4-(x-4)^2}+3\)
y=\(2^{x-4}+3\)
edit: Yes, i was missing the y=, thank you Melody
Ah, i forgot about the 'y=' as i was getting myself confused with solving it as Eq1=Eq2, and i forgot i needed the y= at the beginning. Thank you
+118658 Well the first thing I did was check that they do intersect.
I did this by hand but here Desmos has done it.
ok so there are 2 points.
\(y=\sqrt{4-(x-4)^2}+3\\ y=2^{x-4}+3\\~\\ \sqrt{4-(x-4)^2}+3=2^{x-4}+3\\ \sqrt{4-(x-4)^2}=2^{x-4}\\ \sqrt{-x^2+8x-12}=2^{x-4}\\ 2^{4-x}\sqrt{-x^2+8x-12}=1\\ 2^{4-x}\sqrt{-x^2+8x-12}-1=0\\ \)
I don't know how to solve that ....
Here is what Wolfram|Alpha has to say about it.
http://www.wolframalpha.com/input/?i=sqrt%7B4-(x-4)%5E2%7D%2B3%3D2%5E(x-4)%2B3
That is it from me. 
