Question

find equations of all lines having slope - 3 that are tangent to the curve ( y=\frac{12}{x - 5} ).
select the correct choice below and fill in the answer box(es) within your choice.
a. there are two lines tangent to the curve with a slope of - 3. the equation of the line with the larger y - intercept is ( square ) and the equation of the line with the smaller y - intercept is ( square ).
(type equations.)
b. there is only one line tangent to the curve with a slope of - 3 and its equation is ( square ).
(type an equation.)

Explanation:

Step1: Find the derivative of the function

The function is \(y = \frac{12}{x - 5}=(x - 5)^{-12}\). Using the power rule \((u^n)^\prime=nu^{n - 1}u^\prime\), where \(u=x - 5\), \(n=- 12\) and \(u^\prime = 1\). The derivative \(y^\prime=-\frac{12}{(x - 5)^2}\).

Step2: Set the derivative equal to the slope

We want the slope \(y^\prime=-3\). So, \(-\frac{12}{(x - 5)^2}=-3\). Cross - multiply gives \(12 = 3(x - 5)^2\). Then \((x - 5)^2 = 4\). Taking square roots, \(x-5=\pm2\).

• Case 1: \(x-5 = 2\), then \(x = 7\)
• Case 2: \(x-5=-2\), then \(x = 3\)

Step3: Find the corresponding \(y\) - values

• When \(x = 7\), \(y=\frac{12}{7 - 5}=6\)
• When \(x = 3\), \(y=\frac{12}{3 - 5}=-6\)

Step4: Use the point - slope form \(y - y_1=m(x - x_1)\)

• For the point \((7,6)\) and \(m=-3\), \(y-6=-3(x - 7)\), which simplifies to \(y=-3x + 27\)
• For the point \((3,-6)\) and \(m=-3\), \(y+6=-3(x - 3)\), which simplifies to \(y=-3x+3\)

Answer:

A. There are two lines tangent to the curve with a slope of \(-3\). The equation of the line with the larger \(y\) - intercept is \(y=-3x + 27\) and the equation of the line with the smaller \(y\) - intercept is \(y=-3x+3\)