Look at the graph below carefully
Observe the results of shifting ={2}^{x}f(x)=2x
vertically:
The domain, (−∞,∞) remains unchanged.
When the function is shifted up 3 units to ={2}^{x}+3g(x)=2x +3:
The y-intercept shifts up 3 units to (0,4).
The asymptote shifts up 3 units to y=3y=3.
The range becomes (3,∞).
When the function is shifted down 3 units to ={2}^{x}-3h(x)=2 x −3:
The y-intercept shifts down 3 units to (0,−2).
The asymptote also shifts down 3 units to y=-3y=−3.
The range becomes (−3,∞).
.3125 expressed as a fraction is 5/16
Answer:
Ummm ok?
Step-by-step explanation:
Where’s the question?
An equation without exponents and two variables, is typically a straight line. All the points on the line with integer coordinates are solutions of the equation. Since x and y have to be positive as well, there aren't that many solutions.
Let's see where the line crosses the x-axis, it is where y=0:
x/0.5 + 0 = 18, so x=9 at the intercept. y=0 there, so this is a point on the line, but not a solution to the question (y was supposed to be positive).
Possible values for x are thus limited to 1,2,3,4,5,6 and 7. You can try them all (ie., solve the equation with them) and see for which x values the y is also positive and integer.
You will find that x=4, y=2 is the only pair that satisfies these conditions.
