Math Problem Statement

For each function, find (a) the equation of the secant line through the points where x has the given values, and (b) the equation of the tangent line when x has the first value. Function: f(x) = x^2 + 2x; x1 = 3, x2 = 5.

Solution

Let's break down the problem based on the image.

The problem asks you to:

  1. Find the equation of the secant line through the points where xx has the given values.
  2. Find the equation of the tangent line when xx has the first value.

The function provided is:

f(x)=x2+2xf(x) = x^2 + 2x

Given values for xx:

  • x1=3x_1 = 3
  • x2=5x_2 = 5

Step 1: Equation of the Secant Line

The slope of the secant line between two points x1x_1 and x2x_2 is calculated by: msecant=f(x2)f(x1)x2x1m_{\text{secant}} = \frac{f(x_2) - f(x_1)}{x_2 - x_1}

First, let's calculate the function values at these points:

  • f(3)=32+2(3)=9+6=15f(3) = 3^2 + 2(3) = 9 + 6 = 15
  • f(5)=52+2(5)=25+10=35f(5) = 5^2 + 2(5) = 25 + 10 = 35

Now, the slope of the secant line: msecant=351553=202=10m_{\text{secant}} = \frac{35 - 15}{5 - 3} = \frac{20}{2} = 10

The equation of the secant line is: yf(x1)=msecant(xx1)y - f(x_1) = m_{\text{secant}}(x - x_1) Substitute msecant=10m_{\text{secant}} = 10, f(x1)=15f(x_1) = 15, and x1=3x_1 = 3: y15=10(x3)y - 15 = 10(x - 3) Simplifying: y=10x30+15=10x15y = 10x - 30 + 15 = 10x - 15

Step 2: Equation of the Tangent Line

The slope of the tangent line is the derivative of f(x)f(x) evaluated at x1=3x_1 = 3.

The derivative of f(x)=x2+2xf(x) = x^2 + 2x is: f(x)=2x+2f'(x) = 2x + 2

Evaluating at x=3x = 3: f(3)=2(3)+2=6+2=8f'(3) = 2(3) + 2 = 6 + 2 = 8

The equation of the tangent line is: yf(x1)=f(x1)(xx1)y - f(x_1) = f'(x_1)(x - x_1) Substitute f(x1)=8f'(x_1) = 8, f(x1)=15f(x_1) = 15, and x1=3x_1 = 3: y15=8(x3)y - 15 = 8(x - 3) Simplifying: y=8x24+15=8x9y = 8x - 24 + 15 = 8x - 9

Final Results:

  1. Secant line equation: y=10x15y = 10x - 15
  2. Tangent line equation: y=8x9y = 8x - 9

Would you like more details on this, or do you have any specific questions?

Here are 5 related questions to explore:

  1. How do we find the derivative of a function?
  2. Why is the slope of the tangent line different from the secant line?
  3. What are the geometric interpretations of secant and tangent lines?
  4. How would these equations change if the function were f(x)=x3+xf(x) = x^3 + x?
  5. Can we find the secant line using another pair of points?

Tip: The tangent line always represents the instantaneous rate of change of the function at a specific point.

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Math Problem Analysis

Mathematical Concepts

Algebra
Calculus
Derivatives
Secant Line
Tangent Line

Formulas

Secant line slope formula: m_secant = (f(x2) - f(x1)) / (x2 - x1)
Equation of a line: y - y1 = m(x - x1)
Derivative of a function: f'(x)
Tangent line slope at a point: f'(x1)

Theorems

Definition of Derivative
Point-Slope Form of a Line

Suitable Grade Level

Grades 11-12 or early college