a. i Find the graph in the attachment
ii. The vertical asymptotes are at x = 2 and x = 3
iii. we find the vertical asysmtotes by finding the L.C.M of the expression and equating to zero.
b. i. At x = - 1 there is a critical point.
ii. there isn't a vertical asymptote at x = - 1 because it also has a critical point at x = -1.
c. The value of x when y = 0 is x = ±4.796
a i Graph the original equation
The graph of the equation is in the attachment
ii. What are the vertical asymptotes?
From the graph, the vertical asymptotes are at x = 2 and x = 3
So, the vertical asymptotes are at x = 2 and x = 3
iii. How we can use the equation to verify the asymptotes?
Since y = (x + 5)/(x + 1) ÷ (x + 3)(x - 2)/(x - 4)(x + 1) - 1/(x - 2), we find the vertical asysmtotes by finding the L.C.M of the expression and equating to zero.
So, we find the vertical asysmtotes by finding the L.C.M of the expression and equating to zero.
b. i. What graphical feature occures at x = -1?
The graphical feature that occurs at x = - 1 there is a critical point.
So, at x = - 1 there is a critical point.
ii. Why isn't there a vertical asymptote at x = - 1?
There isn't a vertical asymptote at x = - 1 because
- the factor (x + 1) cancels out in both numerator and denominator
- and also, from the graph, the function is undefined at x = - 1 and
- it also has a critical point at x = -1.
So, there isn't a vertical asymptote at x = - 1 because it also has a critical point at x = -1.
c. Solve the equation for x when y = 0
Since y = (x + 5)/(x + 1) ÷ (x + 3)(x - 2)/(x - 4)(x + 1) - 1/(x - 2), when y = 0, we have
0 = (x + 5)/(x + 1) ÷ (x + 3)(x - 2)/(x - 4)(x + 1) - 1/(x - 2)
(x + 5)(x - 4)(x + 1)/(x + 1)(x + 3)(x - 2) = 1/(x - 2)
(x + 5)(x - 4)/(x + 3)(x - 2) = 1/(x - 2)
(x + 5)(x - 4)/(x + 3) = (x - 2)/(x - 2)
(x + 5)(x - 4)/(x + 3) = 1
(x + 5)(x - 4) = (x + 3)
x² - 4x + 5x - 20 = x + 3
x² + x - 20 = x + 3
x² + x - x - 20 - 3 = 0
x² + 0 - 23 = 0
x² - 23 = 0
x² = 23
x = ±√23
x = ±4.796
So, the value of x when y = 0 is x = ±4.796
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