find equations of all lines having slope - 3 that are tangent to the curve $y = \\frac{12}{x - 5}$. select…

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.)
Answer
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)