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 and the (type equations.) b. there is only one line tangent to the curve with a slope of - 3 and its equation is. (type an equation.)

Explanation:

Step1: Find the derivative of the function

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

Step2: Set the derivative equal to the slope

Since the slope of the tangent line is \(-3\), we set \(y^\prime=-3\).
So, \(-\frac{12}{(x - 5)^2}=-3\).
Cross - multiply: \(12 = 3(x - 5)^2\).
Divide both sides by \(3\): \((x - 5)^2 = 4\).
Take square roots: \(x-5=\pm2\).
When \(x - 5 = 2\), \(x=7\); when \(x - 5=-2\), \(x = 3\).

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 \((x_1 = 7,y_1 = 6)\) and \(m=-3\), \(y - 6=-3(x - 7)\).
Expand: \(y-6=-3x + 21\), so \(y=-3x+27\).
For the point \((x_1 = 3,y_1=-6)\) and \(m=-3\), \(y+6=-3(x - 3)\).
Expand: \(y + 6=-3x+9\), so \(y=-3x + 3\).

Answer:

A. The equation of the line with the larger \(y\) - intercept is \(y=-3x + 27\) and the other is \(y=-3x+3\)