1. Suppose y varies inversely with x, and y = 3 when x = 8.
a. Write an equation that models the variation.
b. Find y when x = 72.
2. What is the quotient in simplest form? State any restrictions on the variable.
(x^2 - 9)/(x+2) / (x-3)/(x^2 + 5x + 6)
3. Describe the graph of y = 2/(x+5) - 3 compared to the graph of y = 1/x.
Hint: Think of transformations of functions.
1.
a. The equation that models the inverse variation is y = k/x, where k is the constant of variation. To find k, we can use the given values:
3 = k/8
k = 24
Therefore, the equation that models the variation is y = 24/x.
b. To find y when x = 72, we plug in x = 72 into the equation:
y = 24/72 = 1/3
2.
To simplify the quotient, we first need to factorize the denominators:
(x^2 - 9) = (x - 3)(x + 3)
(x^2 + 5x + 6) = (x + 3)(x + 2)
So, the expression becomes:
[(x - 3)(x + 3)]/(x+2) / (x-3)/(x + 3)(x + 2)
Now, we can simplify this further by dividing:
[(x - 3)(x + 3)]/(x+2) * (x + 3)(x + 2)/(x - 3)
= (x + 3)/(x + 2)
Restrictions on the variable: The restrictions are x cannot be equal to -2 and 3, as these values would result in division by zero.
3. The graph of y = 2/(x+5) - 3 compared to the graph of y = 1/x involves two transformations of a basic hyperbola y = 1/x.
The graph of y = 1/x is a hyperbola that passes through the points (1, 1), (2, 0.5), (3, 0.33), etc.
When we add 5 to x in the function 2/(x+5), it shifts the graph of 1/x 5 units to the left.
Subtracting 3 from the whole function y = 2/(x+5) shifts the graph 3 units down.
Therefore, the graph of y = 2/(x+5) - 3 is a hyperbola shifted 5 units to the left and 3 units down compared to the graph of y = 1/x.