Q2+ORQ+Freeman+McBrayer+Sahigian

√ 2x x-4 We know that the first equation not have a domain of all real numbers, because x cannot be a negative number with y still being a real number (there is no real number solution to the square root of a negative number).
 * __Formulas__**

__**Graph**__ //√2x////√2(2)=y 2-4=y // //(√2x)^2=(x-4)^2 √4=y -2=y// //2x=x^2-8x+16 2=y// //-2x -2x// //0=x^2-2x-8x+16 √2x=y x-4=y// //0=x^2-10x+16 √2(8)=y 8-4=y// //0=(x-2)(x-8) √16=y 4=y// //0=x-2 0=x-8 4=y// //+2 +2 +8 +8// //x=2 x=8//
 * __Intercept __**

//2 or **__8__** - x=2 doesn't produce the same results for both equations, y is either 2 or -2 meaning that the correct x value is 8.// //(8, 4) is the point of intersection.// //√2(2)=4 but 2-4=-2// //√2(8)=4 and 8-4=4//

__** Solution**__ x>8 (8,+∞) The answer was determined using the intersection found in part 5. 8 was the x-value in the point of intersection. Any x-values above 8 would satisfy the inequality because the answer is when x-4 is greater than // √2x and x-4 is greater than //// √2x past the intersection at 8. Thus the answer would be x>8. //