ORQ+Q2+Evan+Vasu+Cristina

1. y=9/x y=5x+3

2. The functions we used were 9/x and 5x+3. The function of 5x+3 will have a domain of all real numbers, however the function 9/x will never hit zero, therefore it does not have a domain of all real numbers. The functions will intersect at two points, as shown on the graph. Since the function is a linear function, 5x+3 increases at a constant rate and it intercepts t he inverse function of 9/x. The points of these interceptions are (1.0748, 8.3739) and (-1.6748, -5.3739). 3. 9/x ≥ 5x+3

4. (ignore the x on the 9/x line)

5. 9/x ≥ 5x+3 x(9/x) ≥ (5x+3)x 9 ≥ 5x²+3x 0 ≥ 5x²+3x-9 0 = 5x²+3x-9

x = (-b±√(b²-4ac)/(2a) x = (-3±√((3)²-4(5)(-9))/(2(5)) x = (-3±√(9-(-180))/(10) x = (-3±√(189)/(10) x = (-3±√(9) √ (21))/(10) x = (-3±3√ (21))/(10) x = -.3±.3√ (21)

x = -.3+.3√ (21) or 1.0748 x = -.3-.3√ (21) or -1.6748

y = 5(-.3+.3√ (21))+3 y = -1.5+1.5√ (21)+3 y = 1.5+1.5√ (21)

y = 5(-.3-.3√ (21))+3 y = -1.5-1.5√ (21)+3 y = 1.5-1.5√ (21)

y = 1.5+1.5√ (21) or 8.3739 y = 1.5-1.5√ (21) or -5.3739

6. Intersection points: (-.3+.3√ (21), 1.5+1.5√ (21)) or (1.0748, 8.3739) (-.3-.3√ (21), 1.5-1.5√ (21)) or (-1.6748, -5.3739)

Solution: Symbol: x≤ -.3-.3√ (21) or 0<x≤ -.3+.3√ (21) Interval Notation: (-∞,-.3-.3√ (21)] U (0, -.3+.3√ (21)]

7. To determine where the two functions intersect, I had to set the two functions equal to each other, 9/x = 5x+3. Then the goal was to form a function equal to zero. To do so, I first multiplied each side by x, so it was 9 = 5x²+3x, and then i subtracted 9 from both sides to form a quadratic function equal to 0, 0=5x²+3x-9. Next I had to find the x-intercepts, to determine the x values where the original functions intersected. Since I could not use factoring on the quadratic function, i had to use the quadratic formula, x = (-b±√(b²-4ac)/(2a). Once i found x, i put the value back in to the original equations to determine the corresponding y values of their intersection points. Once I found the intersection points, I then looked to the graph s o see when 9/x is greater than or equal to 5x+3. With both the intersection points and the graph i was able to determine which x values corresponded to when 9/x was greater than or equal to 5x+3. I also had to remember that when x=0 could not be included into the answer because when x=0 9/x does not exist on the graph.The solution included all values of x less than and equal to the intersection in the III quadrant, and all values of x greater than 0 and less than and equal to the intersection point in the first quadrant.