Math Problem Statement

Consider the curve y = (x + 1) / (x - 7). Find the slope m of the tangent line at the point (6, -7). Find the equation of the tangent line to the curve at the point (6, -7).

Solution

The given curve is:

y=x+1x7y = \frac{x + 1}{x - 7}

You are asked to:

  1. Find the slope mm of the tangent line at the point (6,7)(6, -7).
  2. Find the equation of the tangent line at that point.

Step 1: Find the slope mm

To find the slope of the tangent line, you need the derivative of the function. The function y=x+1x7y = \frac{x + 1}{x - 7} is a quotient, so we will apply the quotient rule.

The quotient rule states: (uv)=vuuvv2\left( \frac{u}{v} \right)' = \frac{v \cdot u' - u \cdot v'}{v^2}

For this problem:

  • u(x)=x+1u(x) = x + 1
  • v(x)=x7v(x) = x - 7

Now, compute the derivatives:

  • u(x)=1u'(x) = 1
  • v(x)=1v'(x) = 1

Using the quotient rule:

= \frac{x - 7 - (x + 1)}{(x - 7)^2} = \frac{x - 7 - x - 1}{(x - 7)^2} = \frac{-8}{(x - 7)^2}$$ Thus, the derivative is: $$y' = \frac{-8}{(x - 7)^2}$$ To find the slope at the point $$(6, -7)$$, plug $$x = 6$$ into the derivative: $$y'(6) = \frac{-8}{(6 - 7)^2} = \frac{-8}{(-1)^2} = -8$$ So, the slope of the tangent line at $$x = 6$$ is: $$m = -8$$ ### Step 2: Find the equation of the tangent line The equation of the tangent line is given by the point-slope form: $$y - y_1 = m(x - x_1)$$ where $$m$$ is the slope, and $$(x_1, y_1)$$ is the point of tangency. Here, $$m = -8$$, $$x_1 = 6$$, and $$y_1 = -7$$. Substitute the values: $$y - (-7) = -8(x - 6)$$ $$y + 7 = -8(x - 6)$$ Simplify: $$y + 7 = -8x + 48$$ $$y = -8x + 48 - 7$$ $$y = -8x + 41$$ ### Final Answer: 1. The slope $$m = -8$$. 2. The equation of the tangent line is $$y = -8x + 41$$. Let me know if you'd like more details or have any questions! Here are some related questions: 1. How do you apply the quotient rule in other situations? 2. What happens if you compute the derivative using a different method, like product rule? 3. How would the equation change if the curve was shifted horizontally? 4. What if the point of tangency changes to another point? 5. How do you check if the tangent line actually touches the curve only at one point? **Tip:** When using the quotient rule, carefully compute the numerator and denominator separately to avoid mistakes.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Tangent Lines
Quotient Rule

Formulas

Quotient Rule: (u/v)' = (v * u' - u * v') / v^2
Point-Slope Form: y - y1 = m(x - x1)

Theorems

Quotient Rule

Suitable Grade Level

Grades 11-12 (Advanced High School Calculus)