Math Problem Statement

Find the equation of the tangent line at (2, f(2)) when f(2) = 13 and f'(2) = 4.

Solution

To find the equation of the tangent line at the point (2,f(2))(2, f(2)) given f(2)=13f(2) = 13 and f(2)=4f'(2) = 4, we use the point-slope form of a line:

yy1=m(xx1)y - y_1 = m (x - x_1)

where:

  • (x1,y1)(x_1, y_1) is the point on the curve.
  • mm is the slope of the tangent line, which is given by the derivative f(x)f'(x) at x=x1x = x_1.

Given Values:

  • Point on the curve: (x1,y1)=(2,f(2))=(2,13)(x_1, y_1) = (2, f(2)) = (2, 13)
  • Slope of the tangent line: m=f(2)=4m = f'(2) = 4

Plugging in the Values:

Using the point-slope form:

y13=4(x2)y - 13 = 4(x - 2)

To express this equation in standard form or slope-intercept form, we can simplify further:

y13=4x8y - 13 = 4x - 8 y=4x8+13y = 4x - 8 + 13 y=4x+5y = 4x + 5

Final Equation of the Tangent Line:

y=4x+5y = 4x + 5

Would you like more details or have any questions?

Here are 5 related questions you might find helpful:

  1. How do you find the slope of a tangent line at a given point?
  2. What is the difference between the tangent line and the secant line?
  3. How can you find the equation of a normal line to a curve?
  4. What role does the derivative play in determining the slope of a tangent?
  5. How does the point-slope form of a line compare to the slope-intercept form?

Tip: The point-slope form is particularly useful when you know a point on a line and the slope. It is often used to write the equation of tangent lines to functions.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Calculus
Tangent Line
Derivatives

Formulas

Point-slope form of a line: y - y1 = m(x - x1)
Derivative as slope of tangent

Theorems

Differentiation

Suitable Grade Level

College level calculus or advanced high school