Math Problem Statement

Find the equation of the tangent line to the graph of the function f(x)=x+3/x−10 at the point (0,−(3/10)).

Enter the exact answer.

Solution

To find the equation of the tangent line to the graph of the function f(x)=x+3x10f(x) = \frac{x + 3}{x - 10} at the point (0,310)(0, -\frac{3}{10}), follow these steps:

Step 1: Find the derivative of f(x)f(x)

The slope of the tangent line at any point on the curve is given by the derivative of the function at that point. To differentiate f(x)=x+3x10f(x) = \frac{x + 3}{x - 10}, we can apply the quotient rule.

The quotient rule is given by: (u(x)v(x))=v(x)u(x)u(x)v(x)(v(x))2\left( \frac{u(x)}{v(x)} \right)' = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{(v(x))^2} where:

  • u(x)=x+3u(x) = x + 3
  • v(x)=x10v(x) = x - 10

Now, compute the derivatives of u(x)u(x) and v(x)v(x):

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

Applying the quotient rule: f(x)=(x10)(1)(x+3)(1)(x10)2f'(x) = \frac{(x - 10)(1) - (x + 3)(1)}{(x - 10)^2} Simplifying the numerator: f(x)=x10x3(x10)2=13(x10)2f'(x) = \frac{x - 10 - x - 3}{(x - 10)^2} = \frac{-13}{(x - 10)^2}

Step 2: Find the slope of the tangent line at x=0x = 0

To find the slope of the tangent line at the point (0,310)(0, -\frac{3}{10}), substitute x=0x = 0 into the derivative: f(0)=13(010)2=13100f'(0) = \frac{-13}{(0 - 10)^2} = \frac{-13}{100}

So, the slope of the tangent line at x=0x = 0 is m=13100m = -\frac{13}{100}.

Step 3: Use the point-slope form to find the equation of the tangent line

The point-slope form of the equation of a line is: yy1=m(xx1)y - y_1 = m(x - x_1) where mm is the slope, and (x1,y1)(x_1, y_1) is a point on the line.

We know:

  • m=13100m = -\frac{13}{100}
  • (x1,y1)=(0,310)(x_1, y_1) = (0, -\frac{3}{10})

Substitute these values into the point-slope form: y(310)=13100(x0)y - \left( -\frac{3}{10} \right) = -\frac{13}{100}(x - 0) Simplifying: y+310=13100xy + \frac{3}{10} = -\frac{13}{100}x Now, solve for yy: y=13100x310y = -\frac{13}{100}x - \frac{3}{10}

Final Answer:

The equation of the tangent line is: y=13100x310y = -\frac{13}{100}x - \frac{3}{10}

Would you like any further details or clarifications on this?

Here are five related questions:

  1. How do you apply the quotient rule for more complicated functions?
  2. What is the point-slope form of the equation of a line and when should it be used?
  3. How would you find the equation of the tangent line if the function was given in implicit form?
  4. Can you differentiate rational functions using the product rule instead of the quotient rule?
  5. How do you find the slope of a tangent line for trigonometric functions?

Tip: Always double-check the calculations when applying the quotient rule—sign errors can happen, especially when simplifying terms.

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Math Problem Analysis

Mathematical Concepts

Differentiation
Tangent Line
Quotient Rule
Point-Slope Form

Formulas

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

Theorems

Quotient Rule
Point-Slope Form of a Line

Suitable Grade Level

Grades 11-12